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>
long long z = 1000000007;
using namespace std;
long long n, m, ans = 0;
long long arr[300005][8];
pair<long long, long long> g;
long long l, r, mid;
void check() {
  if (l > r) return;
  int flag = 0;
  long long val[n];
  for (long long i = 0; i < n; i++) val[i] = 0;
  long long curr = 1;
  for (long long i = 0; i < m; i++) {
    for (long long j = 0; j < n; j++) {
      if (arr[j][i] >= mid) {
        val[j] += curr;
      }
    }
    curr *= 2;
  }
  map<long long, long long> ma;
  for (long long i = 0; i < n; i++) {
    ma[val[i]] = 1;
  }
  long long upp = pow(2, m);
  long long ans_1 = 0, ans_2 = 0;
  for (long long i = 0; i < upp; i++) {
    for (long long j = 0; j < upp; j++) {
      if ((ma[i] == 1) && (ma[j] == 1) && ((i | j) == (upp - 1))) {
        flag = 1;
        ans_1 = i;
        ans_2 = j;
        break;
      }
    }
    if (flag) break;
  }
  if (flag == 0) {
    r = mid - 1;
    mid = (l + r) / 2;
    check();
  } else {
    for (long long i = 0; i < n; i++) {
      if (val[i] == ans_1) {
        g.second = i;
      }
      if (val[i] == ans_2) g.first = i;
    }
    l = mid + 1;
    mid = (l + r) / 2;
    check();
  }
  return;
}
int main() {
  cin >> n >> m;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < m; j++) {
      cin >> arr[i][j];
    }
  }
  l = 0, r = 1000000000;
  mid = (l + r) / 2;
  check();
  cout << g.first + 1 << ' ' << g.second + 1 << endl;
  return 0;
}