codeforce868c

C. Qualification Rounds

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Snark and Philip are preparing the problemset for the upcoming pre-qualification round for semi-quarter-finals. They have a bank of nproblems, and they want to select any non-empty subset of it as a problemset.

k experienced teams are participating in the contest. Some of these teams already know some of the problems. To make the contest interesting for them, each of the teams should know at most half of the selected problems.

Determine if Snark and Philip can make an interesting problemset!

Input

The first line contains two integers n, k (1 ≤ n ≤ 105, 1 ≤ k ≤ 4) — the number of problems and the number of experienced teams.

Each of the next n lines contains k integers, each equal to 0 or 1. The j-th number in the i-th line is 1 if j-th team knows i-th problem and 0 otherwise.

Output

Print "YES" (quotes for clarity), if it is possible to make an interesting problemset, and "NO" otherwise.

You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES").

Examples

input

5 3
1 0 1
1 1 0
1 0 0
1 0 0
1 0 0

output

NO

input

3 2
1 0
1 1
0 1

output

YES

Note

In the first example you can't make any interesting problemset, because the first team knows all problems.

In the second example you can choose the first and the third problems.

 

题意：n个题，m个人，从n个题中挑k道题，使m个人每个人会的题目数不超过k/2。

输入的意思是：第i行第j列为1，说明第j个人会第i道题，为0说明不会

结论：k直接取2就行   取大了反倒不好满足

首先我们先分成两种情况

k ＝ 1       就看有没有全0行嘛，对吧

k >= 2      首先考虑p道题满足的情况下的，首先p道题我们就确定这已经是一个稳定的系统了，那么我们再加一行，这一行肯定有至少一个1（否则这道题全是0，就是第一种情况了），p加一行的话肯定某一个人会多对一道题，意思就是，其它人的正确率会下降，但是不会导致p＋1这个集合变得不符合题意，但是加1的那一列却有可能导致集合不符合题意。

所以，直接取k ＝ 2即可，在k＝2上面加题反倒可能会使集合可能坏掉

 

其次说一下具体实现，一共就最多4个人，一道题，最多也就是2的4次方，16种排列。直接枚举每一种即可，

用一个ans[16]表示每种情况出现过几次，在输入的时候就能够处理

 

附上代码

#include <iostream>
using namespace std;
int arr[16] = {0};
int main()
{
    int n,m,i,j,t,temp;
    cin >> n >> m;
    for(i = 0; i < n; ++i)
    {
        temp = 0;
        for(j = 0; j < m; ++j)
        {
            scanf("%d",&t);
            temp |= t<<j;
        }
        arr[temp]++;
    }
    for(i = 0;i < 16; ++i)
    {
        for(j = 0; j < 16; ++j)
        {
            if((i&j) == 0 && arr[i] && arr[j])
            {
                cout << "YES" << endl;
                return 0;
            }
        }
    }
    cout << "NO" << endl;
}