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;
using ll = long long;
using ul = unsigned long long;
using pii = pair<int, int>;
using vvi = vector<vector<int>>;
using vi = vector<int>;
using vl = vector<ll>;
using pll = pair<ll, ll>;
using vsi = vector<set<int>>;
ll power(ll a, ll n) {
  ll j = 0;
  ll x = 1;
  while (1ll << j < n) {
    if (1ll << j & n) x = (x * a) % 1000000007;
    a = a * a % 1000000007;
  }
  return x;
}
int main() {
  int n, m;
  cin >> n >> m;
  int a[n][m];
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) cin >> a[i][j];
  }
  int l = 0;
  int r = 1e9;
  int mask = 0;
  for (int i = 0; i < m; i++) mask += 1 << i;
  int ani, anj;
  while (l <= r) {
    int mid = (l + r) / 2;
    vvi isp(1 << m);
    bool check = false;
    for (int i = 0; i < n; i++) {
      int temp = 0;
      for (int j = 0; j < m; j++) {
        if (a[i][j] >= mid) temp += 1 << j;
      }
      isp[temp].push_back(i);
    }
    int ai, aj;
    for (int i = 0; i < 1 << m; i++) {
      for (int j = i; j < 1 << m; j++) {
        if (isp[j].size() && isp[i].size() && ((i | j) == mask)) {
          check = true;
          ai = isp[i][0];
          aj = isp[j][0];
        }
      }
    }
    if (check) {
      ani = ai;
      anj = aj;
      l = mid + 1;
    } else
      r = mid - 1;
  }
  cout << ani + 1 << ' ' << anj + 1 << endl;
}