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;
void solution();
template <class T>
T GCD(T a, T b) {
  if (a == 0) return b;
  return GCD(b % a, a);
}
template <class T>
T modpow(T x, T p) {
  T res = 1;
  while (p > 0) {
    if (p & 1) res *= x;
    x *= x;
    p /= 2;
  }
  return res;
}
template <class T>
void print(T a[], long long n) {
  for (long long i = 0; i < n; i++) std::cout << a[i] << ' ';
  cout << endl;
}
template <class T>
void printc(T c) {
  for (auto const& v : c) std::cout << v << ' ';
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solution();
  return 0;
}
int mod = 1e9 + 7;
int dp[50][2000];
int n, m;
void bottomup() {
  memset(dp, 0, sizeof dp);
  dp[0][0] = 1;
  for (int i = 1; i <= m + m; i++) {
    for (int j = 1; j <= n; j++) {
      for (int k = 0; k <= j; k++) {
        dp[i][j] += dp[i - 1][k];
        if (dp[i][j] >= mod) dp[i][j] -= mod;
      }
    }
  }
  int total = accumulate(dp[m + m], dp[m + m] + n + 1, 0ll) % mod;
  cout << total << endl;
}
int a1 = 0, a2 = 0;
vector<vector<int>> a;
;
bool can(int mid) {
  vector<int> msk(1 << m, -1);
  for (int i = 0; i < n; i++) {
    int cur = 0;
    for (int j = 0; j < m; j++) {
      if (a[i][j] >= mid) {
        cur ^= (1 << j);
      }
    }
    msk[cur] = i;
  }
  if (msk[(1 << m) - 1] != -1) {
    a1 = msk[(1 << m) - 1];
    a2 = msk[(1 << m) - 1];
    return true;
  }
  for (int i = 0; (i < (1 << m)); i++) {
    for (int j = 0; (j < (1 << m)); j++) {
      if (msk[i] != -1 && msk[j] != -1 && (i | j) == (1 << m) - 1) {
        a1 = msk[i];
        a2 = msk[j];
        return true;
      }
    }
  }
  return false;
}
void solution() {
  cin >> n >> m;
  a.resize(n, vector<int>(m));
  for (long long i = 0; i < n; ++i) {
    for (long long j = 0; j < m; ++j) {
      cin >> a[i][j];
    }
  }
  int lo = 0, hi = int(1e9) + 43;
  while (lo <= hi) {
    int mid = (lo + hi) / 2;
    if (can(mid))
      lo = mid + 1;
    else
      hi = mid - 1;
  }
  cout << a1 + 1 << " " << a2 + 1 << endl;
}