Codeforces Round #412 (rated, Div. 2, base on VK Cup 2017 Round 3) D - Dynamic Problem Scoring

地址：http://codeforces.com/contest/807/problem/D

题目：

D. Dynamic Problem Scoring

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya and Petya take part in a Codeforces round. The round lasts for two hours and contains five problems.

For this round the dynamic problem scoring is used. If you were lucky not to participate in any Codeforces round with dynamic problem scoring, here is what it means. The maximum point value of the problem depends on the ratio of the number of participants who solved the problem to the total number of round participants. Everyone who made at least one submission is considered to be participating in the round.

Pay attention to the range bounds. For example, if 40 people are taking part in the round, and 10 of them solve a particular problem, then the solvers fraction is equal to 1 / 4, and the problem's maximum point value is equal to 1500.

If the problem's maximum point value is equal to x, then for each whole minute passed from the beginning of the contest to the moment of the participant's correct submission, the participant loses x / 250 points. For example, if the problem's maximum point value is 2000, and the participant submits a correct solution to it 40 minutes into the round, this participant will be awarded with 2000·(1 - 40 / 250) = 1680points for this problem.

There are n participants in the round, including Vasya and Petya. For each participant and each problem, the number of minutes which passed between the beginning of the contest and the submission of this participant to this problem is known. It's also possible that this participant made no submissions to this problem.

With two seconds until the end of the round, all participants' submissions have passed pretests, and not a single hack attempt has been made. Vasya believes that no more submissions or hack attempts will be made in the remaining two seconds, and every submission will pass the system testing.

Unfortunately, Vasya is a cheater. He has registered 109 + 7 new accounts for the round. Now Vasya can submit any of his solutions from these new accounts in order to change the maximum point values of the problems. Vasya can also submit any wrong solutions to any problems. Note that Vasya can not submit correct solutions to the problems he hasn't solved.

Vasya seeks to score strictly more points than Petya in the current round. Vasya has already prepared the scripts which allow to obfuscate his solutions and submit them into the system from any of the new accounts in just fractions of seconds. However, Vasya doesn't want to make his cheating too obvious, so he wants to achieve his goal while making submissions from the smallest possible number of new accounts.

Find the smallest number of new accounts Vasya needs in order to beat Petya (provided that Vasya's assumptions are correct), or report that Vasya can't achieve his goal.

Input

The first line contains a single integer n (2 ≤ n ≤ 120) — the number of round participants, including Vasya and Petya.

Each of the next n lines contains five integers ai, 1, ai, 2..., ai, 5 ( - 1 ≤ ai, j ≤ 119) — the number of minutes passed between the beginning of the round and the submission of problem j by participant i, or -1 if participant i hasn't solved problem j.

It is guaranteed that each participant has made at least one successful submission.

Vasya is listed as participant number 1, Petya is listed as participant number 2, all the other participants are listed in no particular order.

Output

Output a single integer — the number of new accounts Vasya needs to beat Petya, or -1 if Vasya can't achieve his goal.

Examples

input

2
5 15 40 70 115
50 45 40 30 15

output

2

input

3
55 80 10 -1 -1
15 -1 79 60 -1
42 -1 13 -1 -1

output

3

input

5
119 119 119 119 119
0 0 0 0 -1
20 65 12 73 77
78 112 22 23 11
1 78 60 111 62

output

27

input

4
-1 20 40 77 119
30 10 73 50 107
21 29 -1 64 98
117 65 -1 -1 -1

output

-1

Note

In the first example, Vasya's optimal strategy is to submit the solutions to the last three problems from two new accounts. In this case the first two problems will have the maximum point value of 1000, while the last three problems will have the maximum point value of 500. Vasya's score will be equal to 980 + 940 + 420 + 360 + 270 = 2970 points, while Petya will score just 800 + 820 + 420 + 440 + 470 = 2950 points.

In the second example, Vasya has to make a single unsuccessful submission to any problem from two new accounts, and a single successful submission to the first problem from the third new account. In this case, the maximum point values of the problems will be equal to 500, 1500, 1000, 1500, 3000. Vasya will score 2370 points, while Petya will score just 2294 points.

In the third example, Vasya can achieve his goal by submitting the solutions to the first four problems from 27 new accounts. The maximum point values of the problems will be equal to 500, 500, 500, 500, 2000. Thanks to the high cost of the fifth problem, Vasya will manage to beat Petya who solved the first four problems very quickly, but couldn't solve the fifth one.

思路：

　　先考虑极限情况需要的人数：把一个120人都过了的题目的分值改成3000分题需要3720人。

　　所以所需最多人数在4000以内。

　　然后暴力从小到大枚举人数个数，对于每个人数x，Vasya都采用对自己最有利的策略。

　　ps1：这题不能二分，不具有二分性质，人数多反而可能会对自己不利。

　　ps2：每题得分是整数，所以不要用double存储，用int

　　

#include <bits/stdc++.h>

using namespace std;

#define MP make_pair
#define PB push_back
typedef long long LL;
typedef pair<int,int> PII;
const double eps=1e-8;
const double pi=acos(-1.0);
const int K=1e6+7;
const int mod=1e9+7;

int n,ans=-1,a[200][10],sum[10];

int sc(int k,int x)
{
    double ta,tb;
    if(a[2][k]==-1&&a[1][k]!=-1)
        tb=1.0-a[1][k]/250.0;
    else if(a[1][k]==-1&&a[2][k]!=-1)
        tb=-1.0+a[2][k]/250.0;
    else
        tb=(a[2][k]-a[1][k])/250.0;
    if(tb>0||a[1][k]==-1)
       ta=sum[k]*1.0/(n+x);
   else
       ta=(sum[k]+x)*1.0/(n+x);
   if(ta>0.5)
        return 500*tb;
   else if(ta>0.25)
        return 1000*tb;
   else if(ta>0.125)
        return 1500*tb;
   else if(ta>pow(2.0,-4.0))
        return 2000*tb;
   else if(ta>pow(2.0,-5.0))
        return 2500*tb;
   else
        return 3000*tb;
}
bool check(int x)
{
    int ret=0;
    for(int i=1;i<=5;i++)
        ret+=sc(i,x);
    return ret>0;
}
int main(void)
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    for(int j=1;j<=5;j++)
    scanf("%d",&a[i][j]),a[i][j]>=0?sum[j]++:0;
    for(int i=0;i<=4000&&ans==-1;i++)
    if(check(i))
        ans=i;
    printf("%d\n",ans);
    return 0;
}