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 md = 1e9 + 7;
const int xn = 3e5 + 10;
const int xm = 5e5 + 10;
const int SQ = 700;
const long long inf = 1e18 + 10;
int a[xn][8], n, m;
int f[300], ff[300];
bool fl[xn];
pair<int, int> ans;
void bin(int s) {
  int ss = s;
  for (int i = 0; i < 10; i++) fl[i] = 0;
  int t = 0;
  while (s) {
    if (s % 2) fl[t] = 1;
    t++;
    s /= 2;
  }
  int mx = -1, dx = -1;
  for (int i = 0; i < n; i++) {
    int mn = -1;
    for (int j = 0; j < m; j++) {
      if (!fl[j]) continue;
      if (mn == -1) mn = a[i][j];
      mn = min(mn, a[i][j]);
    }
    if (mx == -1 || mx < mn) dx = i + 1;
    mx = max(mx, mn);
  }
  f[ss] = dx;
  ff[ss] = mx;
}
bool check(int x) {
  for (int i = 1; i < pow(2, m) - 1; i++) {
    if (ff[i] >= x && ff[int(pow(2, m) - i - 1)] >= x) {
      ans = {f[i], f[int(pow(2, m) - i - 1)]};
      return true;
    }
  }
  int k = pow(2, m) - 1;
  if (ff[k] >= x) {
    ans = {f[k], f[k]};
    return true;
  }
  return false;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> m;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) cin >> a[i][j];
  for (int i = 0; i < pow(2, m); i++) bin(i);
  int l, r, mid;
  l = 0, r = 1e9 + 10;
  while (r - l > 1) {
    mid = (l + r) / 2;
    if (check(mid))
      l = mid;
    else
      r = mid;
  }
  if (l == 0) {
    cout << 1 << ' ' << 1 << '\n';
    return 0;
  }
  cout << ans.first << ' ' << ans.second << '\n';
  return 0;
}