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;
template <typename T>
void _dbg(const char* _s, T _h) {
  cerr << _s << " = " << _h << "\n";
}
template <typename T, typename... Ts>
void _dbg(const char* _s, T _h, Ts... _t) {
  int _b = 0;
  while (((_b += *_s == '(') -= *_s == ')') != 0 || *_s != ',') cerr << *_s++;
  cerr << " = " << _h << ",";
  _dbg(_s + 1, _t...);
}
const int64_t INF = static_cast<int64_t>(1e9) + 7;
const int64_t LINF = INF * INF;
const int MAXN = static_cast<int>(1e6) + 17;
void solve() {
  int n, m;
  cin >> n >> m;
  int ai = 0, aj = 0;
  int bnd = (1 << m);
  vector<int> nums(m), cur;
  vector<pair<int, int>> vec(bnd, {0, 0});
  int ans = 0;
  for (int i = 0; i < n; ++i) {
    cur.assign(bnd, INF);
    for (int j = 0; j < m; ++j) {
      cin >> nums[j];
    }
    for (int j = 0; j < bnd; ++j) {
      for (int k = 0; k < m; ++k) {
        if (j & (1 << k)) {
          cur[j] = min(cur[j], nums[k]);
        }
      }
    }
    for (int j = 0; j < bnd; ++j) {
      int cand = min(vec[j].first, cur[bnd - 1 - j]);
      if (cand > ans) {
        ans = cand;
        ai = vec[j].second;
        aj = i;
      }
      if (cur[j] > vec[j].first) {
        vec[j].first = cur[j];
        vec[j].second = i;
      }
    }
  }
  cout << ai + 1 << ' ' << aj + 1 << '\n';
}
int main() {
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios_base::sync_with_stdio(false);
  int q = 1;
  for (int i = 0; i < q; ++i) {
    solve();
  }
}