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 MXN = 3e5 + 10;
const long long LOG = 15;
const long long MXM = (1LL << 8);
long long n, m, ans, a, b;
long long A[MXN][LOG], dp[MXM], id[MXM], Min[MXN];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < m; j++) cin >> A[i][j];
  }
  dp[0] = 1e9 + 2;
  for (int mask = 1; mask < (1LL << m); mask++) {
    long long bit = __builtin_ctz(mask);
    dp[mask] = min(dp[mask - (1LL << bit)], A[1][bit]), id[mask] = 1;
  }
  ans = dp[(1LL << m) - 1], a = b = 1;
  for (int i = 2; i <= n; i++) {
    Min[0] = 1e9 + 2;
    for (int mask = 1; mask < (1LL << m); mask++) {
      long long bit = __builtin_ctz(mask);
      Min[mask] = min(Min[mask - (1LL << bit)], A[i][bit]);
    }
    for (int mask = 1; mask < (1LL << m); mask++) {
      long long oth = ((1LL << m) - 1) ^ mask;
      long long now = min(Min[mask], dp[oth]);
      ans = max(ans, now);
      if (ans == now) {
        if (mask < (1LL << m) - 1)
          a = i, b = id[oth];
        else
          a = b = i;
      }
    }
    for (int mask = 1; mask < (1LL << m); mask++) {
      dp[mask] = max(dp[mask], Min[mask]);
      if (dp[mask] == Min[mask]) id[mask] = i;
    }
  }
  cout << a << ' ' << b << '\n';
  return 0;
}