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 long long INF = (long long)0x3f3f3f3f3f3f3f, MAX = 9e18, MIN = -9e18;
const int N = 1e6 + 10, M = 2e6 + 10, mod = 1e9 + 7, inf = 0x3f3f3f;
long long a[N][10], v[N];
long long n, m, ans1, ans2, mx = 0, r;
bool check(long long x) {
  long long total = (1 << m);
  for (int i = 1; (i) <= (total); i++) v[i] = 0;
  for (int i = 1; (i) <= (n); i++) {
    int tmp = 0;
    for (int j = 0; (j) < (m); j++)
      if (a[i][j] >= x) tmp |= (1 << j);
    v[tmp] = i;
  }
  for (int i = 0; (i) < (total); i++)
    for (int j = 0; (j) < (total); j++)
      if (v[i] && v[j] && (i | j) == total - 1) {
        if (x > mx) mx = x, ans1 = v[i], ans2 = v[j];
        return 1;
      }
  return 0;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> m;
  for (int i = 1; (i) <= (n); i++) {
    long long tmp = MAX;
    for (int j = 0; (j) < (m); j++)
      cin >> a[i][j], tmp = min(tmp, a[i][j]), r = max(r, a[i][j]);
    if (tmp > mx) ans1 = ans2 = i, mx = tmp;
  }
  long long l = mx;
  r++;
  while (l < r) {
    long long mid = l + r >> 1;
    if (check(mid))
      l = mid + 1;
    else
      r = mid;
  }
  if (!ans1) ans1 = 1;
  if (!ans2) ans2 = 1;
  cout << ans1 << ' ' << ans2 << '\n';
  return 0;
}