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 N = 3 * 1e5, M = 8;
int n, m;
int a[N + 5][M + 5];
int mx, ansx, ansy;
int tmpx, tmpy;
int b[N + 5], d[(1 << M) + 5];
inline int read() {
  int x = 0, ff = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') ff = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  return x * ff;
}
bool able(int x) {
  memset(b, 0, sizeof(b));
  memset(d, 0, sizeof(d));
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) {
      if (a[i][j] >= x) b[i] |= (1 << j);
      d[b[i]] = i + 1;
    }
  for (int i = 0; i < n; i++)
    for (int j = 1; j < (1 << m); j++)
      if (d[j] && (b[i] | j) == (1 << m) - 1) {
        tmpx = i + 1;
        tmpy = d[j];
        return 1;
      }
  return 0;
}
int main() {
  n = read(), m = read();
  int l = 0, r = 0, mid;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) {
      a[i][j] = read();
      r = max(a[i][j], r);
    }
  while (l <= r) {
    mid = (l + r) / 2;
    if (able(mid)) {
      mx = mid;
      ansx = tmpx, ansy = tmpy;
      l = mid + 1;
    } else
      r = mid - 1;
  }
  if (mx > 0)
    cout << ansx << " " << ansy << endl;
  else
    cout << "1 1\n";
  return 0;
}