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;
template <class T>
inline bool uin(T& a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
inline bool uax(T& a, T b) {
  return a < b ? (a = b, true) : false;
}
const int nax = 3e5 + 10;
int x = -1, y = -1;
int A[nax][8];
int N, M;
bool check(int val) {
  vector<int> id(1 << M, -1);
  for (int i = 0; i < N; ++i) {
    int mask = 0;
    for (int j = 0; j < M; ++j)
      if (A[i][j] >= val) mask += (1 << j);
    id[mask] = i;
  }
  if (id[(1 << M) - 1] != -1) {
    x = y = id[(1 << M) - 1];
    return true;
  }
  for (int i = 0; i < (1 << M); ++i) {
    for (int j = i + 1; j < (1 << M); ++j) {
      if (id[i] != -1 && id[j] != -1 && ((i | j) == (1 << M) - 1)) {
        x = id[i], y = id[j];
        return true;
      }
    }
  }
  return false;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> N >> M;
  for (int i = 0; i < N; ++i)
    for (int j = 0; j < M; ++j) cin >> A[i][j];
  int l = -1, r = 1e9;
  while (l < r) {
    int m = (l + r + 1) >> 1;
    if (check(m))
      l = m;
    else
      r = m - 1;
  }
  cout << x + 1 << ' ' << y + 1 << '\n';
}