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;
int n, m, a[300005][10], l, r, md, arr[10000007], mn, jog, ans, nw, b, bit;
pair<int, int> dp[1 << 9];
bitset<9> bset;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      scanf("%d", &a[i][j]);
      arr[++nw] = a[i][j];
    }
  }
  for (int i = 1; i <= (1 << m); i++) {
    for (int j = 1; j <= n; j++) {
      mn = 1e9;
      for (int k = 1; k <= m; k++)
        if ((i & (1 << (k - 1)))) mn = min(a[j][k], mn);
      if (dp[i].first < mn) {
        dp[i].first = mn;
        dp[i].second = j;
      }
    }
  }
  sort(arr + 1, arr + 1 + nw);
  l = 1, r = nw;
  while (l <= r) {
    md = (l + r) / 2;
    b = 0;
    for (int i = 1; i <= n; i++) {
      bit = 0;
      for (int j = 1; j <= m; j++) {
        if (a[i][j] < arr[md]) {
          bit |= (1 << (j - 1));
        }
      }
      if (bit == 0) {
        b = 1;
        ans = i, jog = i;
      } else if (dp[bit].first >= arr[md]) {
        b = 1;
        ans = i, jog = dp[bit].second;
      }
    }
    if (b == 1)
      l = md + 1;
    else
      r = md - 1;
  }
  printf("%d %d", ans, jog);
}