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 LIM = 1e5 + 5, MOD = 1e9 + 7;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long fastpowMOD(long long a, long long p, long long MOD) {
  if (p == 0) return 1;
  long long z = fastpowMOD(a, p / 2, MOD);
  z = (z * z) % MOD;
  if (p % 2) z = (z * a) % MOD;
  return z;
}
bool seive[100005];
long long dp[10005][105];
void SieveOfEratosthenes(long long n) {
  memset(seive, true, sizeof(seive));
  for (long long p = 2; p * p <= n; p++)
    if (seive[p] == true)
      for (long long i = p * p; i <= n; i += p) seive[i] = false;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, i, j, k, m;
  cin >> n >> m;
  long long l = 0;
  long long r = 1e9;
  long long a[n + 5][m + 5];
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      cin >> a[i][j];
    }
  }
  long long o = 3 | 5;
  while (l < r) {
    long long mid = (l + r + 1) / 2;
    long long val[257] = {0};
    for (i = 0; i < n; i++) {
      long long sum = 0;
      for (j = 0; j < m; j++) {
        if (a[i][j] >= mid) {
          sum += (1 << j);
        }
      }
      val[sum]++;
    }
    long long flag = 0;
    for (i = 0; i <= 255; i++) {
      for (j = 0; j <= 255; j++) {
        if ((i | j) == (1 << m) - 1) {
          if (val[i] and val[j]) {
            flag = 1;
          }
        }
      }
    }
    if (flag) {
      l = mid;
    } else {
      r = mid - 1;
    }
  }
  long long mid = l;
  long long val[257] = {0};
  for (i = 0; i < n; i++) {
    long long sum = 0;
    for (j = 0; j < m; j++) {
      if (a[i][j] >= mid) {
        sum += (1 << j);
      }
    }
    val[sum] = i + 1;
  }
  long long flag = 0;
  long long c1, c2;
  for (i = 0; i <= 255; i++) {
    for (j = 0; j <= 255; j++) {
      if ((i | j) == (1 << m) - 1) {
        if (val[i] and val[j]) {
          flag = 1;
          c1 = val[i];
          c2 = val[j];
        }
      }
    }
  }
  cout << c1 << " " << c2;
}