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 ≤ i, j ≤ 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 ∈ [1, m] b_k = max(a_{i, k}, a_{j, k}).

Your goal is to choose i and j so that the value of min _{k = 1}^{m} b_k is maximum possible.

Input

The first line contains two integers n and m (1 ≤ n ≤ 3 ⋅ 10^5, 1 ≤ m ≤ 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 ≤ a_{x, y} ≤ 10^9).

Output

Print two integers i and j (1 ≤ i, j ≤ n, it is possible that i = j) — the indices of the two arrays you have to choose so that the value of min _{k = 1}^{m} 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e5 + 10;
const int inf = 0x3f3f3f3f;
const long long mod = 1e9 + 7;
const long long linf = 0x3f3f3f3f3f3f3f3f;
int n, m, arr[maxn][10], len, brr[maxn];
bool solve(int v) {
  bool vis[520] = {0};
  for (int i = 1; i <= n; i++) {
    int now = 0;
    for (int j = 1; j <= m; j++) {
      now <<= 1;
      if (arr[i][j] >= v) now += 1;
    }
    if (vis[now]) return 1;
    now ^= len;
    if (!vis[now]) {
      for (int k = now; k <= len; k++) {
        if ((k & now) == now) vis[k] = 1;
      }
    }
  }
  return 0;
}
int main() {
  scanf("%d%d", &n, &m);
  if (n == 1) {
    cout << 1 << ' ' << 1;
    return 0;
  }
  len = (1 << m) - 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) scanf("%d", &arr[i][j]);
  }
  int l = 0, r = 1e9 + 10;
  while (l < r - 1) {
    int mid = l + r >> 1;
    if (solve(mid))
      l = mid;
    else
      r = mid;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      brr[i] <<= 1;
      if (arr[i][j] >= l) {
        brr[i] += 1;
      }
    }
  }
  bool vis[520] = {0};
  int a, b;
  for (int i = 1; i <= n; i++) {
    if (vis[brr[i]]) {
      b = i;
      break;
    }
    int now = len ^ brr[i];
    if (!vis[now]) {
      for (int k = now; k <= len; k++) {
        if ((k & now) == now) vis[k] = 1;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    if ((brr[i] | brr[b]) == len) {
      cout << i << ' ' << b;
      break;
    }
  }
  return 0;
}