Codeforces Round #316 (Div. 2) A 水

A. Elections

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

The country of Byalechinsk is running elections involving n candidates. The country consists of m cities. We know how many people in each city voted for each candidate.

The electoral system in the country is pretty unusual. At the first stage of elections the votes are counted for each city: it is assumed that in each city won the candidate who got the highest number of votes in this city, and if several candidates got the maximum number of votes, then the winner is the one with a smaller index.

At the second stage of elections the winner is determined by the same principle over the cities: the winner of the elections is the candidate who won in the maximum number of cities, and among those who got the maximum number of cities the winner is the one with a smaller index.

Determine who will win the elections.

Input

The first line of the input contains two integers n, m (1 ≤ n, m ≤ 100) — the number of candidates and of cities, respectively.

Each of the next m lines contains n non-negative integers, the j-th number in the i-th line a_ij (1 ≤ j ≤ n, 1 ≤ i ≤ m, 0 ≤ a_ij ≤ 10⁹) denotes the number of votes for candidate j in city i.

It is guaranteed that the total number of people in all the cities does not exceed 10⁹.

Output

Print a single number — the index of the candidate who won the elections. The candidates are indexed starting from one.

Examples

Input

3 3
1 2 3
2 3 1
1 2 1

Output

2

Input

3 4
10 10 3
5 1 6
2 2 2
1 5 7

Output

1

Note

Note to the first sample test. At the first stage city 1 chosen candidate 3, city 2 chosen candidate 2, city 3 chosen candidate 2. The winner is candidate 2, he gained 2 votes.

Note to the second sample test. At the first stage in city 1 candidates 1 and 2 got the same maximum number of votes, but candidate 1 has a smaller index, so the city chose candidate 1. City 2 chosen candidate 3. City 3 chosen candidate 1, due to the fact that everyone has the same number of votes, and 1 has the smallest index. City 4 chosen the candidate 3. On the second stage the same number of cities chose candidates 1 and 3. The winner is candidate 1, the one with the smaller index.

题意：投票选州长  m个城市 n个候选人    给你n个候选人在每个城市的得票数   在当前城市获票最多的候选人获胜 当票数一样时 编号小的

候选人获胜  统计获得城市获胜最多的候选人作为州长  当出现获胜次数相同的情况下 输出编号最小的一名

题解： 水

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<queue>
 #include<stack>
 #include<vector>
 #include<map>
 #include<algorithm>
 #define ll __int64
 #define mod 1e9+7
 #define PI acos(-1.0)
 using namespace std;
 int n,m;
 int exm;
 int biao;
 map<int,int> mp;
 int main()
 {
     int ans=;
     int maaa=;
     scanf("%d %d",&n,&m);
     mp.clear();
     for(int i=;i<=m;i++)
     {
         int biao=;
         int maxn=-;
         for(int j=;j<=n;j++)
         {
             scanf("%d",&exm);
             if(exm>maxn)
             {
                 maxn=exm;
                 biao=j;
             }
         }
         mp[biao]++;
         if(mp[biao]>maaa||(mp[biao]==maaa&&biao<ans))
         {
             maaa=mp[biao];
             ans=biao;
         }
     }
     cout<<ans<<endl;
     return ;
 }