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 t, d, ct, q, ans, l, m, r, n, mx, mn, i, j, k, x, y, z, s;
char ch;
string ss = "", str = "";
int a[300005][9], b[300005];
void dbg(int x, int y = 0, int z = 0, int a = 0) {
  cout << "Dbg: ";
  cout << x << " " << y << " " << z << " " << a << "\n";
}
bool ask(int k) {
  vector<pair<int, int>> v;
  bool vis[260] = {0};
  for (int i = 0; i < n; i++) {
    int no = 0;
    for (int j = 0; j < m; j++) {
      if (a[i][j] >= k) no += (1 << j);
    }
    if (!vis[no]) vis[no] = 1, v.push_back({no, i});
  }
  for (int i = 0; i < v.size(); i++)
    for (int j = i; j < v.size(); j++)
      if ((v[i].first | v[j].first) == ((1 << m) - 1)) {
        x = v[i].second;
        y = v[j].second;
        return 1;
      }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  cin >> n >> m;
  for ((i) = 0; (i) < n; (i)++)
    for ((j) = 0; (j) < m; (j)++) cin >> a[i][j];
  l = 0;
  r = 1000000000;
  while (l <= r) {
    int mid = (l + r) / 2;
    if (ask(mid))
      l = mid + 1, ans = mid;
    else
      r = mid - 1;
  }
  cout << x + 1 << " " << y + 1;
  return 0;
}