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 int INF = 2e9;
const long long BIG_INF = 1e18;
const long long MOD = 1e9 + 7;
const int maxn = 3e5 + 5, maxm = 9;
int a[maxn][maxm], n, m, max_mask;
vector<pair<int, int> > masks, unique_masks;
pair<int, int> get_inds(int x) {
  masks.clear();
  masks.resize(n);
  unique_masks.clear();
  for (int i = 0; i < n; i++) {
    int mask = 0;
    for (int j = 0; j < m; j++)
      if (a[i][j] >= x) mask |= 1 << j;
    masks[i] = {mask, i};
  }
  sort(masks.begin(), masks.end());
  int last_ind = 0;
  for (int i = 1; i < n; i++) {
    if (masks[i].first == masks[last_ind].first) continue;
    unique_masks.push_back(masks[last_ind]);
    last_ind = i;
  }
  unique_masks.push_back(masks[last_ind]);
  for (int i = 0; i < (int)unique_masks.size(); i++)
    for (int j = 0; j < (int)unique_masks.size(); j++)
      if ((unique_masks[i].first | unique_masks[j].first) == max_mask)
        return {unique_masks[i].second, unique_masks[j].second};
  return {-1, -1};
}
bool prov(int x) {
  pair<int, int> res = get_inds(x);
  return res.first != -1 && res.second != -1;
}
int bin_search() {
  int l = 0, r = 1e9 + 1;
  while (r - l > 1) {
    int m = (l + r) / 2;
    if (prov(m))
      l = m;
    else
      r = m;
  }
  return l;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) cin >> a[i][j];
  max_mask = (1 << m) - 1;
  pair<int, int> ans = get_inds(bin_search());
  cout << ans.first + 1 << " " << ans.second + 1 << "\n";
  return 0;
}