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 MOD = 1e9 + 7;
const long long ARR_MAX = 3e5 + 7;
const long long VAL_MAX = 256;
const long long INF = 1e10 + 1;
long long a[ARR_MAX][9];
long long n, m;
pair<bool, pair<long long, long long> > check_ans(long long mid) {
  long long sz = 1ll << m;
  map<long long, long long> ind;
  pair<bool, pair<long long, long long> > res;
  for (int i = 0; i < n; i++) {
    long long val = 0, bit = 1;
    for (int j = 0; j < m; j++) {
      if (a[i][j] >= mid) {
        val += bit;
      }
      bit *= 2;
    }
    ind[val] = i;
  }
  res.second.first = -1;
  res.second.second = -1;
  res.first = 0;
  for (auto it = ind.begin(); it != ind.end(); it++) {
    for (auto it1 = ind.begin(); it1 != ind.end(); it1++) {
      if ((it->first | it1->first) == (sz - 1)) {
        res.first = 1;
        res.second.first = it->second;
        res.second.second = it1->second;
        break;
      }
    }
    if (res.first) break;
  }
  return res;
}
void solve() {
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> a[i][j];
    }
  }
  long long lo = 0, hi = INF, ans_x = -1, ans_y = -1;
  while (lo < hi) {
    long long mid = (lo + hi + 1) / 2;
    pair<bool, pair<long long, long long> > ans = check_ans(mid);
    if (ans.first) {
      lo = mid;
      ans_x = ans.second.first;
      ans_y = ans.second.second;
    } else {
      hi = mid - 1;
    }
  }
  pair<bool, pair<long long, long long> > ans = check_ans(lo);
  cout << ans.second.first + 1 << " " << ans.second.second + 1, cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}