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;
mt19937 rnd(time(0));
bool isPrime(long long a) {
  long long i;
  for (i = 2; i * i <= a; i++) {
    if (a % i == 0) {
      return 0;
    }
  }
  return (a != 1);
}
bool isPalindrom(const string& s, int l, int r) {
  for (int i = 0; i < (r - l + 1) / 2; i++) {
    if (s[l + i] != s[r - i]) {
      return 0;
    }
  }
  return 1;
}
template <class type1>
type1 gcd(type1 a, type1 b) {
  type1 t;
  while (b) {
    t = b;
    b = a % b;
    a = t;
  }
  return a;
}
long long positiveMod(long long v, long long md) {
  v = v % md;
  if (v < 0) {
    v += md;
  }
  return v;
}
vector<int> dat[300005];
int bestVal[600], bestInd[600];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  for (auto& it : bestVal) {
    it = -1;
  }
  for (auto& it : bestInd) {
    it = -1;
  }
  int n, m, i, j, a, b, c, k, l, r, sz;
  cin >> n >> m;
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      cin >> a;
      dat[i].push_back(a);
    }
  }
  for (l = 1; l < (1 << m); l++) {
    vector<int> ind;
    for (r = 0; r < m; r++) {
      if ((l & (1 << r)) != 0) {
        ind.push_back(r);
      }
    }
    sz = ind.size();
    for (i = 0; i < n; i++) {
      a = (1000000007ll);
      for (j = 0; j < sz; j++) {
        a = min(dat[i][ind[j]], a);
      }
      if (a > bestVal[l]) {
        bestVal[l] = a;
        bestInd[l] = i + 1;
      }
    }
  }
  int res = 0;
  for (l = 1; l < ((1 << m) - 1); l++) {
    res = max(res, min(bestVal[l], bestVal[(l ^ ((1 << m) - 1))]));
  }
  if (bestVal[(1 << m) - 1] >= res) {
    cout << bestInd[(1 << m) - 1] << " " << bestInd[(1 << m) - 1] << "\n";
    return 0;
  }
  for (l = 1; l < ((1 << m) - 1); l++) {
    if (min(bestVal[l], bestVal[(l ^ ((1 << m) - 1))]) == res) {
      cout << bestInd[l] << " " << bestInd[(l ^ ((1 << m) - 1))] << "\n";
      return 0;
    }
  }
  return 0;
}