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>
const int MAXN = 3e5 + 59;
const int MOD = 1e9 + 7;
const int INF = 0x3F3F3F3F;
const long long llINF = 0x3F3F3F3F3F3F3F3F;
using namespace std;
int a[MAXN][8];
int n, m;
int trans(int id, int comp) {
  int x = 0;
  for (int i = m; i >= 1; --i) {
    x = (x << 1) + (a[id][i] >= comp);
  }
  return x;
}
vector<int> chk;
pair<int, int> ans;
bool chek(int cmp) {
  chk = vector<int>(256, -1);
  for (int i = n; i >= 1; --i) {
    chk[trans(i, cmp)] = i;
  }
  for (int i = 0; i < 256; ++i) {
    for (int j = 0; j < 256; ++j) {
      if ((i | j) == (1 << m) - 1 && chk[i] != -1 && chk[j] != -1) {
        ans = {chk[i], chk[j]};
        return true;
      }
    }
  }
  return false;
}
void solve(int kaseId = -1) {
  cin >> n >> m;
  vector<int> vals;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      cin >> a[i][j];
      vals.emplace_back(a[i][j]);
    }
  }
  sort(vals.begin(), vals.end());
  vals.erase(unique(vals.begin(), vals.end()), vals.end());
  int l = 0, r = vals.size() - 1;
  int mid;
  while (l <= r) {
    mid = (l + r) >> 1;
    if (chek(vals[mid])) {
      l = mid + 1;
    } else {
      r = mid - 1;
    }
  }
  cout << ans.first << " " << ans.second << '\n';
}
void solves() {
  int Kase = 0;
  cin >> Kase;
  for (int kase = 1; kase <= Kase; kase++) {
    solve(kase);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  solve();
  return 0;
}