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 maxn = 3e5 + 10;
const long long MOD = 1e9 + 7;
const long long INF = 1e18;
inline long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
inline int lb(int x) { return x & -x; }
long long qpow(long long base, int n) {
  if (base == 0) return 0;
  long long ans = 1;
  while (n) {
    if (n & 1) ans *= base;
    n >>= 1;
    base *= base;
  }
  return ans;
}
int n, m;
long long a[maxn][9];
long long lo = 0, hi = 1e10;
int x, y;
int ve[(1 << 9) - 1], pos[(1 << 9) - 1];
bool solve(long long v) {
  memset(ve, 0, sizeof(ve));
  memset(pos, 0, sizeof(pos));
  for (int i = (1); i <= (n); ++i) {
    int tmp = 0;
    for (int j = (1); j <= (m); ++j)
      if (v <= a[i][j]) tmp |= (1 << (j - 1));
    ve[tmp] = 1;
    pos[tmp] = i;
  }
  int sum = (1 << (m)) - 1;
  for (int i = (0); i <= (sum); ++i) {
    if (!pos[i]) continue;
    for (int j = (i); j <= (sum); ++j)
      if (pos[j] && (i | j) == sum) {
        x = pos[i];
        y = pos[j];
        return true;
      }
  }
  return false;
}
int main() {
  cin >> n >> m;
  for (int i = (1); i <= (n); ++i)
    for (int j = (1); j <= (m); ++j) cin >> a[i][j];
  while (lo < hi) {
    long long mi = (lo + hi) >> 1;
    if (solve(mi))
      lo = mi + 1;
    else
      hi = mi;
  }
  if (x > y) swap(x, y);
  cout << x << ' ' << y << '\n';
}