Minimax Problem

You are given nnn arrays a1,a2,...,ana_1, a_2, ..., a_na1,a2,...,an; each array consists of exactly mmm integers. We denote the yyy-th element of the xxx-th array as ax,ya_{x,y}ax,y.

You have to choose two arrays aia_iai and aja_jaj (1≤i,j≤n1≤i,j≤n1≤i,j≤n, it is possible that i=ji=ji=j). After that, you will obtain a new array bbb consisting of m integers, such that for every k∈[1,m]k∈[1,m]k∈[1,m] bk=max(ai,k,aj,kb_k=max(a_{i,k},a_{j,k}bk=max(ai,k,aj,k).

Your goal is to choose iii and jjj so that the value of min⁡k=1mbk\min \limits_{k = 1}^{m} b_kk=1minmbk is maximum possible.

Input

The first line contains two integers nnn and mmm (1≤n≤3⋅105,1≤m≤81≤n≤3⋅10^5, 1≤m≤81≤n≤3⋅105,1≤m≤8) — the number of arrays and the number of elements in each array, respectively.

Then nnn lines follow, the xxx-th line contains the array axa_xax represented by mmm integers ax,1,ax,2,...,ax,ma_{x,1}, a_{x,2}, ..., a_{x,m}ax,1,ax,2,...,ax,m (0≤ax,y≤1090≤ax,y≤1090≤ax,y≤109).

Output

Print two integers iii and jjj (1≤i,j≤n1≤i,j≤n1≤i,j≤n, it is possible that i=ji=ji=j) — the indices of the two arrays you have to choose so that the value of min⁡k=1mbk\min \limits_{k = 1}^{m} b_kk=1minmbk is maximum possible. If there are multiple answers, print any of them.

Example

input

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;
}

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