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;
template <typename T>
void UpdateMax(T& a, T b) {
  if (a < b) a = b;
}
template <typename T>
void UpdateMin(T& a, T b) {
  if (a > b) a = b;
}
template <typename T>
inline void read(T& x) {
  x = 0;
  char c = getchar();
  int flag = 1;
  while (!isdigit(c)) {
    if (c == '-') flag = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = x * 10 + c - '0';
    c = getchar();
  }
  x *= flag;
}
template <typename T, typename V>
void Print(const pair<T, V>& x) {
  cout << x.first << " @ " << x.second << " | ";
}
template <typename T>
void Print(const T& x) {
  cout << x << " ";
}
template <typename T>
void Pcon(const T& x) {
  cout << "\nsize: " << x.size() << endl;
  for (const auto& item : x) Print(item);
}
const int N = 3e5 + 5;
int n, m, Min, Max, ans1, ans2;
int vis[256];
int a[N][8];
bool Check(int mid) {
  memset(vis, 0, sizeof(vis));
  for (int i = 1; i <= n; i++) {
    int had = 0, notHad = 0;
    for (int j = 1; j <= m; j++) {
      if (a[i][j] >= mid)
        had |= 1 << (j - 1);
      else
        notHad |= 1 << (j - 1);
    }
    if (vis[had]) continue;
    if (!notHad) {
      ans1 = ans2 = i;
      return true;
    }
    if (vis[notHad]) {
      ans1 = vis[notHad];
      ans2 = i;
      return true;
    }
    for (int mask = had; mask; mask = (mask - 1) & had) vis[mask] = i;
  }
  return false;
}
int main() {
  cin.sync_with_stdio(false);
  Min = 1000000000, Max = 0;
  read(n);
  read(m);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) {
      read(a[i][j]);
      UpdateMax(Max, a[i][j]);
      UpdateMin(Min, a[i][j]);
    }
  int left = Min, right = Max;
  while (left != right) {
    int mid = left + (right - left) / 2 + 1;
    if (Check(mid))
      left = mid;
    else
      right = mid - 1;
  }
  Check(left);
  cout << ans1 << " " << ans2 << endl;
}