cf 570 A. Elections

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 aij(1 ≤ j ≤ n, 1 ≤ i ≤ m, 0 ≤ aij ≤ 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.

Sample test(s)

input

output

input

output

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.

给定选举信息，对于同一个城市，得票数最多的人（若相同，则编号最小）的获胜．

对于参选人，获胜城市最多的任获胜（相同，则编号最小）

扫一遍即可

/*************************************************************************
    > File Name: code/cf/#316/A.cpp
    > Author: 111qqz
    > Email: rkz2013@126.com 
    > Created Time: 2015年08月14日 星期五 01时20分40秒
 ************************************************************************/

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#define y0 abc111qqz
#define y1 hust111qqz
#define yn hez111qqz
#define j1 cute111qqz
#define tm crazy111qqz
#define lr dying111qqz
using namespace std;
#define REP(i, n) for (int i=0;i<int(n);++i)  
typedef long long LL;
typedef unsigned long long ULL;
const int inf = 0x7fffffff;
const int N=1E2+5;
int a[N][N];
int n,m;
int mx;
int cnt[N];
int main()
{
    cin>>n>>m;
    for ( int i = 1 ; i<= m ;i ++)
    {
    for ( int j = 1 ; j <= n ; j++ )
    {
        scanf("%d",&a[i][j]);
    }
    }
    memset(cnt,0,sizeof(cnt));
    for ( int i = 1 ;i <= m ; i++ )
    {
    mx = -1;
    int p = -1;
    for ( int j = 1 ; j <= n ; j++ )
    {
        if (a[i][j]>mx)
        {
        mx = a[i][j];
        p = j;
        }
    }
    cnt[p]++;
    }
    mx = -1;
    int ans;
    for ( int i = 1 ; i<= n ; i++)
    {
    if (cnt[i]>mx)
    {
        mx = cnt[i];
        ans = i;
    }
    }
    cout<<ans<<endl;


  
    return 0;
}