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 — 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 cicibytecoins 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 nn and mm (1≤n,m≤30001≤n,m≤3000) — the number of voters and the number of parties respectively.

Each of the following nn lines contains two integers pipi and cici (1≤pi≤m1≤pi≤m, 1≤ci≤1091≤ci≤109) — the index of this voter'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

1 2
1 100

output

0

input

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

output

500

input

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

output

600

題目大概：

有n個投票人，m個候選人，票數最多的候選人獲勝，你想要讓1號候選人獲勝，你需要買通其他投票人，使得他們投給1號票。每個人買通的費用不同，問最少可以多少錢，使得1號獲勝。

思路：

1號獲勝，一定是1號得到k個人投票，其他候選人都少於k個人投票。

在這裡我們可以列舉這個k，貪心的計算1號獲勝的最少花費。

程式碼：

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=4000;
ll INF=1e18+7;
const int mod=1e9+7;
vector<pair <int,int> >a,G[maxn];
bool vis[maxn];

int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    int w,id;
    int sum=0;
    for(int i=1; i<=n; i++)
    {
        scanf("%d%d",&id,&w);
        if(id==1)sum++;
        else
        {
            a.push_back(make_pair(w,i));
            G[id].push_back(make_pair(w,i));
        }
    }
    sort(a.begin(),a.end());

    for(int i=1; i<=m; i++)sort(G[i].begin(),G[i].end());
    ll min_1=INF;

    for(int k=0; k<=n-1; k++)
    {
        memset(vis,0,sizeof(vis));
        ll sum1=0;
        int t=0;
        for(int i=2; i<=m; i++)
        {
            int p=0;
            while(G[i].size()-p>k)
            {
                sum1+=G[i][p].first;
                vis[G[i][p].second]=1;
                p++;
            }
            t+=p;
        }
        int ans=0;
        while(t+sum<=k)
        {
            if(!vis[a[ans].second])
            {
                sum1+=a[ans].first;
                t++;
                vis[a[ans].second]=1;
            }
            ans++;
        }
        min_1=min(min_1,sum1);
    }
    printf("%I64d\n",min_1);
    return 0;
}