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;
void solve();
signed main() {
  clock_t beg = clock();
  cout << fixed << setprecision(15);
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  long long tt = t;
  while (t--) {
    solve();
  }
  cerr << "\nExecution time: " << (clock() - beg) / 1000 << '\n';
  return 0;
}
const long long N = 3e5 + 55;
const long long M = 8;
long long arr[N][M];
long long mask[N];
long long cnt[1 << M];
pair<long long, long long> ans;
long long n, m;
bool check(long long mid) {
  memset(cnt, 0, sizeof cnt);
  for (long long i = 1; i <= n; ++i) {
    mask[i] = 0;
    for (long long bit = 0; bit < m; ++bit) {
      if (arr[i][bit] >= mid) mask[i] += (1ll << bit);
    }
    cnt[mask[i]] = i;
  }
  for (long long msk = 0; msk < (1 << m); ++msk) {
    for (long long msk2 = 0; msk2 < (1 << m); ++msk2) {
      if (cnt[msk] and cnt[msk2] and (msk | msk2) == ((1 << m) - 1)) {
        ans = {cnt[msk], cnt[msk2]};
        return true;
      }
    }
  }
  return false;
}
void solve() {
  cin >> n >> m;
  long long lo = 0, hi = 1e9;
  for (long long i = (1); i < (n + 1); ++i)
    for (long long j = (0); j < (m); ++j) cin >> arr[i][j];
  pair<long long, long long> fina;
  while (lo <= hi) {
    long long mid = (lo + hi) / 2;
    if (check(mid)) {
      fina = ans;
      lo = mid + 1;
    } else {
      hi = mid - 1;
    }
  }
  cout << fina.first << ' ' << fina.second;
}