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;
int n, m;
string ve[500];
int v[300005][10];
int poz[500];
unordered_set<string> us;
int sol, x1, x2;
bool se_pupa(string a, string b) {
  for (int i = 0; i < m; i++)
    if (a[i] + b[i] - '0' - '0' == 0) return false;
  return true;
}
bool valid(int val) {
  us.clear();
  int contor = 0;
  for (int i = 1; i <= n; i++) {
    string str = "";
    for (int j = 1; j <= m; j++) {
      if (v[i][j] >= val) {
        str += '1';
      } else
        str += '0';
    }
    if (us.count(str) == 0) {
      us.insert(str);
      ve[++contor] = str;
      poz[contor] = i;
    }
  }
  for (int i = 1; i <= contor; i++) {
    for (int j = i; j <= contor; j++) {
      if (se_pupa(ve[i], ve[j]) == true) {
        x1 = poz[i];
        x2 = poz[j];
        return true;
      }
    }
  }
  return false;
}
void cb(int st, int dr) {
  if (st > dr) return;
  int mid = (st + dr) / 2;
  if (valid(mid) == true) {
    sol = mid;
    cb(mid + 1, dr);
  } else
    cb(st, mid - 1);
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) cin >> v[i][j];
  }
  cb(0, 1000000000);
  cout << x1 << " " << x2;
  return 0;
}