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;
int arr[300001][8];
int buildMask(int k) {
  int mask = 0;
  for (int x = 0; x < m; x++) {
    if (k & (1 << x)) {
      mask = mask & ~(1 << x);
    } else {
      mask = mask | (1 << x);
    }
  }
  return mask;
}
pair<int, int> check(int v) {
  int combo[(int)pow(2, m) + 1];
  memset(combo, 0, sizeof combo);
  for (int x = 0; x < n; x++) {
    int mask = 0;
    for (int y = 0; y < m; y++) {
      if (arr[x][y] >= v) {
        mask = mask | (1 << y);
        combo[mask] = x + 1;
      }
    }
    int nMask = buildMask(mask);
    if (nMask == 0) return make_pair(x + 1, x + 1);
    int loo = pow(2, m);
    for (int y = 0; y < loo; y++) {
      if (combo[y] && (y & nMask) == nMask) return make_pair(x + 1, combo[y]);
    }
  }
  return make_pair(0, 0);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m;
  for (int x = 0; x < n; x++) {
    for (int y = 0; y < m; y++) cin >> arr[x][y];
  }
  long long lo = 0;
  long long hi = 1e9;
  pair<int, int> ans;
  long long val = -1;
  while (lo <= hi) {
    long long mid = (lo + hi) / 2;
    pair<int, int> cur = check(mid);
    if (cur.first != 0) {
      lo = mid + 1;
      if (mid > val) ans = cur;
      val = max(val, mid);
    } else {
      hi = mid - 1;
    }
  }
  cout << ans.first << " " << ans.second << endl;
}