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 INF = 1e9 + 7;
const int MOD = 1e9 + 7;
int n, m;
int bitmask;
int a[300100][10];
int id1, id2;
bool ok(int val) {
  vector<int> d[300];
  for (int i = 1; i <= n; i++) {
    int tmp = 0;
    for (int j = 1; j <= m; j++)
      if (a[i][j] >= val) tmp += (1 << (j - 1));
    d[tmp].push_back(i);
  }
  for (int i = 0; i < (1 << m); i++)
    for (int j = 0; j < (1 << m); j++)
      if ((i | j) == bitmask) {
        if (d[i].size() > 0 && d[j].size() > 0) {
          id1 = d[i][0];
          id2 = d[j][0];
          return true;
        }
      }
  return false;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> m;
  bitmask = (1 << m) - 1;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) cin >> a[i][j];
  int lo, hi, mid, res;
  lo = 0;
  hi = INF;
  while (lo <= hi) {
    mid = (lo + hi) >> 1;
    if (ok(mid)) {
      res = mid;
      lo = mid + 1;
    } else
      hi = mid - 1;
  }
  cout << id1 << " " << id2;
  return 0;
}