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 N = 100010;
const int MX = 123456;
const int mod = (int)1e9 + 7;
const int base = 97;
const unsigned long long BS1 = 10000019ULL;
const int INF = (1 << 29);
template <class T>
inline void fastScan(T &x) {
  register char c = getchar();
  int neg = 0;
  x = 0;
  for (; (c < 48 || c > 57) && (c != '-'); c = getchar())
    ;
  if (c == '-') {
    neg = 1;
    c = getchar();
  }
  for (; c > 47 && c < 58; c = getchar()) {
    x = (x << 1) + (x << 3) + c - 48;
  }
  if (neg) x = -x;
}
int n, m;
int l, r;
vector<vector<int> > inp;
int main() {
  fastScan(n);
  fastScan(m);
  inp.resize(n, vector<int>(m, 0));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      fastScan(inp[i][j]);
    }
  }
  int low = 0, high = 1e9, ans = 1e9;
  while (low <= high) {
    int mid = (low + high) >> 1;
    int possible = 0;
    vector<int> maskV(256, -1);
    for (int i = 0; i < n && !possible; i++) {
      int mask = 0, mn = 1e9;
      for (int j = 0; j < m; j++) {
        if (inp[i][j] >= mid) {
          mask |= (1 << j);
        }
        mn = min(mn, inp[i][j]);
      }
      maskV[mask] = i;
      if (mn >= mid) {
        possible = 1;
        l = i;
        r = i;
      }
    }
    for (int i = 0; i < 256 && !possible; i++) {
      for (int j = 0; j < 256 && !possible; j++) {
        if (maskV[i] != -1 && maskV[j] != -1) {
          if ((i | j) == (1 << m) - 1) {
            l = maskV[i];
            r = maskV[j];
            possible = 1;
          }
        }
      }
    }
    if (possible) {
      low = mid + 1;
      ans = mid;
    } else {
      high = mid - 1;
    }
  }
  printf("%d %d\n", l + 1, r + 1);
}