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>
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("O3", "omit-frame-pointer", "inline")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native,avx2")
using namespace std;
const long long mod = 1000000007;
mt19937 rng((int)chrono::steady_clock::now().time_since_epoch().count());
const long long mxN = 300010;
const long long mxM = 12;
int N;
int exist[(1 << mxM)];
vector<int> a[mxN];
int main(int argc, char *argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t = 1;
  while (t--) {
    int M;
    cin >> N >> M;
    for (int i = 0; i < N; i++) {
      a[i].resize(M);
      for (int j = 0; j < M; j++) cin >> a[i][j];
    }
    long long lo = 0, hi = 1000000000, res = lo, ia = -1, ib = -1;
    while (lo <= hi) {
      long long md = (lo + hi) >> 1;
      memset(exist, 0, sizeof exist);
      bool can = false;
      int ca = -1, cb = -1;
      for (int i = 0; i < N; i++) {
        long long cur = 0;
        for (int j = 0; j < M; j++)
          if (a[i][j] >= md) cur |= (1 << j);
        exist[cur] = i + 1;
      }
      for (int i = 0; i < (1 << M); i++) {
        for (int j = 0; j < (1 << M); j++) {
          if ((i | j) + 1 == (1 << M)) {
            if (exist[i] && exist[j]) {
              can = true;
              ca = exist[i];
              cb = exist[j];
              break;
            }
          }
        }
      }
      if (can)
        lo = md + 1, res = md, ia = ca, ib = cb;
      else
        hi = md - 1;
    }
    cerr << res << '\n';
    cout << ia << ' ' << ib << '\n';
  }
  return 0;
}