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 long long LINF = 1001002003004005006ll;
int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
int n, m;
long long a[300010][10];
long long can = 0, ansi, ansj;
bool check(long long num) {
  vector<bool> v((1 << m), false);
  for (int i = 0; i < n; i++) {
    int bit = 0;
    for (int j = 0; j < m; j++) {
      if (a[i][j] >= num) {
        bit += (1 << j);
      }
    }
    v[bit] = true;
  }
  vector<bool> vv((1 << m), false);
  for (int i = 0; i < (1 << m); i++) vv[i] = v[i];
  for (int i = 0; i < (1 << m); i++) {
    for (int j = i + 1; j < (1 << m); j++) {
      if (v[i] and v[j]) vv[i | j] = true;
    }
  }
  return (vv[(1 << m) - 1]);
}
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(0);
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> a[i][j];
    }
  }
  long long lw = 0, hi = LINF, ok = 0;
  while (hi - lw > 1) {
    long long mid = (hi + lw) / 2;
    if (check(mid))
      lw = mid;
    else
      hi = mid;
  }
  vector<int> v((1 << m), -1);
  for (int i = 0; i < n; i++) {
    int bit = 0;
    for (int j = 0; j < m; j++) {
      if (a[i][j] >= lw) {
        bit += (1 << j);
      }
    }
    v[bit] = i;
  }
  for (int i = 0; i < (1 << m); i++) {
    for (int j = i; j < (1 << m); j++) {
      if (v[i] >= 0 and v[j] >= 0 and (i | j) == ((1 << m) - 1)) {
        cout << v[i] + 1 << ' ' << v[j] + 1 << endl;
        return 0;
      }
    }
  }
  return 0;
}