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 a[300005][11];
pair<int, int> check(int n, int m, int mid) {
  vector<pair<string, int> > s;
  unordered_map<string, int> m1;
  for (int i = 1; i <= n; i++) {
    string cur;
    for (int j = 1; j <= m; j++) {
      if (a[i][j] >= mid)
        cur += "1";
      else
        cur += "0";
    }
    if (!m1[cur]) s.push_back({cur, i}), m1[cur]++;
  }
  vector<pair<string, int> > v(s.begin(), s.end());
  if (v.size() == 1) {
    for (char &c : v[0].first) {
      if (c == '0') return {-1, -1};
    }
    return {v[0].second, v[0].second};
  }
  for (int i = 0; i < v.size(); i++) {
    for (int j = i + 1; j < v.size(); j++) {
      int fl = 1;
      for (int in = 0; in <= m - 1; in++) {
        if (v[i].first[in] == '0' and v[j].first[in] == '0') {
          fl = 0;
          break;
        }
      }
      if (fl == 1) return {v[i].second, v[j].second};
    }
  }
  return {-1, -1};
}
void solve() {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      cin >> a[i][j];
    }
  }
  int l = 0, r = (int)(1e9 + 7) - 7;
  pair<int, int> ans;
  while (l <= r) {
    int mid = (l + r) / 2;
    pair<int, int> cur = check(n, m, mid);
    if (cur.first != -1)
      ans = cur, l = mid + 1;
    else
      r = mid - 1;
  }
  cout << ans.first << ' ' << ans.second;
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
  }
}