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 long double EPS = 1e-5;
const long double PI = acos(-1);
int mod = (int)998244353;
const int MOD7 = 1000000007;
const int MOD9 = 1000000009;
const long long INF = 8.5e18;
const int inf = 1e9;
mt19937 rnd(std::chrono::steady_clock::now().time_since_epoch().count());
int n, m;
vector<vector<int>> a;
pair<int, int> check(int x) {
  vector<int> id(1 << m, -1);
  for (int i = 0; i < n; ++i) {
    int mask = 0;
    for (int j = 0; j < m; ++j)
      if (a[i][j] >= x) mask |= (1 << j);
    id[mask] = i;
  }
  for (int i = (1 << m) - 1; i >= 0; --i) {
    for (int j = 0; j < m; ++j) {
      if ((i >> j) & 1) continue;
      if (id[i ^ (1 << j)] != -1) id[i] = id[i ^ (1 << j)];
    }
  }
  for (int i = 0; i < n; ++i) {
    int tmask = 0;
    for (int j = 0; j < m; ++j) {
      if (a[i][j] >= x) tmask |= (1 << j);
    }
    tmask ^= ((1 << m) - 1);
    if (id[tmask] == -1) continue;
    return {i, id[tmask]};
  }
  return make_pair(-1, -1);
}
signed main() {
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  srand(time(NULL));
  cin >> n >> m;
  a.resize(n, vector<int>(m));
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < m; ++j) cin >> a[i][j];
  int l = -1, r = 1e9 + 228;
  while (r - l > 1) {
    int m = (l + r) / 2;
    if (check(m).first != -1)
      l = m;
    else
      r = m;
  }
  auto fx = check(l);
  cout << fx.first + 1 << ' ' << fx.second + 1;
}