cf 1020 C

C. Elections

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.

Elections are coming. You know the number of voters and the number of parties — $\mathrm{nn and mm respectively.\; For\; each\; voter\; you\; know\; the\; party\; he\; is\; going\; to\; vote\; for.\; However,\; he\; can\; easily\; change\; his\; vote\; given\; a\; certain\; amount\; of\; money.\; In\; particular,\; if\; you\; give ii-th\; voter cici bytecoins\; you\; can\; ask\; him\; to\; vote\; for\; any\; other\; party\; you\; choose.}$

The United Party of Berland has decided to perform a statistical study — you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.

Input

The first line of input contains two integers $\mathrm{nn and mm (1\le n,m\le 30001\le n,m\le 3000) —\; the\; number\; of\; voters\; and\; the\; number\; of\; parties\; respectively.}$

Each of the following $\mathrm{nn lines\; contains\; two\; integers pipi and cici (1\le pi\le m1\le pi\le m, 1\le ci\le 1091\le ci\le 109) —\; the\; index\; of\; this\; voter\text{'}s\; preferred\; party\; and\; the\; number\; of\; bytecoins\; needed\; for\; him\; to\; reconsider\; his\; decision.}$

The United Party of Berland has the index $11.$

Output

Print a single number — the minimum number of bytecoins needed for The United Party of Berland to win the elections.

Examples

input

Copy

1 2
1 100

output

Copy

0

input

Copy

5 5
2 100
3 200
4 300
5 400
5 900

output

Copy

500

input

Copy

5 5
2 100
3 200
4 300
5 800
5 900

output

Copy

600

Note

In the first sample, The United Party wins the elections even without buying extra votes.

In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties $\mathrm{33, 44 and 55 get\; one\; vote\; and\; party\; number 22 gets\; no\; votes.}$

In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party.

 

 

#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <string>
#include <deque>
using namespace std;
#define ll long long 
#define gep(i,a,b) for(int i=a;i<=b;i++)
#define gepp(i,a,b) for(int i=a;i>=b;i--)    
#define gep1(i,a,b) for(ll i=a;i<=b;i++)
#define gepp1(i,a,b) for(ll i=a;i>=b;i--)
#define N   3009
#define ph  push_back
#define  inf 10000000000000098
vector<int>ve[N];
int n,m;
ll MIN;
ll solve(ll  x){
    vector<int>se;
    ll ret=0,cnt=ve[1].size();//初始值
    gep(i,2,m){
        int num=ve[i].size();
        gep(j,0,num-1){
            if(num-j>=x){//严格大
                ret+=ve[i][j];//必须加到party1上，因为加到别的上面，最后还是加到1上。
                cnt++;
            }
            else{
                se.ph(ve[i][j]);
            }
        }
    }
    if(cnt<x){
        sort(se.begin(),se.end());
        for(int i=0;i<se.size()&&cnt<x;i++){//在从se里找
            ret+=se[i];
            cnt++;
        }    
    }
    //cnt最后一定是>=x的
    return ret;
}
int main()
{   
    int x;
    ll y;
    scanf("%d%d",&n,&m);
    gep(i,1,n){
        scanf("%d%lld",&x,&y);
        ve[x].ph(y);
    }
    gep(i,1,m)
    sort(ve[i].begin(),ve[i].end());
    MIN=inf;
    gep(i,1,n){//枚举查找，不能用二分
        MIN=min(MIN,solve(i));
    }
    printf("%lld\n",MIN);
    return 0;
}