Minimax Problem

You are given $n$ arrays $a_1,a_2,...,a_n$; each array consists of exactly $m$ integers. We denote the $y$-th element of the $x$-th array as $a_{x,y}$.

You have to choose two arrays $a_i$ and $a_j$ ($1\le i,j\le n$, it is possible that $i=j$). After that, you will obtain a new array $b$ consisting of m integers, such that for every $k\in [1,m]$ $b_k=max(a_{i,k},a_{j,k}$).

Your goal is to choose $i$ and $j$ so that the value of $\underset{k=1}{\overset{m}{\min }}{b}_k$ is maximum possible.

Input

The first line contains two integers $n$ and $m$ ($1\le n\le 3\cdot 10^5,1\le m\le 8$) — the number of arrays and the number of elements in each array, respectively.

Then $n$ lines follow, the $x$-th line contains the array $a_x$ represented by $m$ integers $a_{x,1},a_{x,2},...,a_{x,m}$ ($0\le ax,y\le 109$).

Output

Print two integers $i$ and $j$ ($1\le i,j\le n$, it is possible that $i=j$) — the indices of the two arrays you have to choose so that the value of $\underset{k=1}{\overset{m}{\min }}{b}_k$ is maximum possible. If there are multiple answers, print any of them.

Example

input

6 5
5 0 3 1 2
1 8 9 1 3
1 2 3 4 5
9 1 0 3 7
2 3 0 6 3
6 4 1 7 0

output

1 5

最小值最大化问题，绝大概率是二分。但是二分之后的judge环节怎么弄呢？这就是本题的出彩之处。我们将每一个数组中的大于等于mid的位置设置为1，否则就设置为0。这样我们进行或操作，这样出来的就是取最大值操作了。这样如果两两相或，如果各个位上都是1，就代表着当前的mid就是最小值，然后不断的二分，就可以找到最优解了。

#include <bits/stdc++.h>
using namespace std;
const int maxn=3e5+10;
int a[maxn][10];
int n,m,x,y;
int vis[390];
int check(int mid) {
    
    
    memset(vis, 0, sizeof(vis));
    for (int i = 1; i <= n; i++) {
    
    
        int k = 0;
        for (int j = 0; j < m; j++) {
    
    
            if (a[i][j] >= mid) k |= (1 << j);
        }
        vis[k] = i;
    }
    for (int i = 0; i <= 255; i++) {
    
    
        for (int j = 0; j <= 255; j++) {
    
    
            if (vis[i] && vis[j] && (i | j) == (1 << m) - 1) {
    
    
                x = vis[i];
                y = vis[j];
                return 1;
            }
        }
    }
    return 0;
}
int main() {
    
    
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) {
    
    
        for (int j = 0; j < m; j++) {
    
    
            scanf("%d", &a[i][j]);
        }
    }
    int l = 0, r = 1e9;
    while (l <= r) {
    
    
        int mid = l + r >> 1;
        if (check(mid)) l = mid + 1; else r = mid - 1;
    }
    printf("%d %d\n", x, y);
    return 0;
}