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>
using vector2d = vector<vector<T>>;
string to_string(string s) { return '"' + s + '"'; }
string to_string(const char* s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto& x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
int n, m;
vector2d<int> a;
pair<int, int> Check(int val) {
  vector<int> index(1 << m, -1);
  for (int i = 0; i < n; i++) {
    int mask = 0;
    for (int j = 0; j < m; j++) {
      if (a[i][j] >= val) {
        mask |= (1 << j);
      }
    }
    index[mask] = i;
  }
  pair<int, int> res = {-1, -1};
  if (index[(1 << m) - 1] != -1) {
    res.first = index[(1 << m) - 1];
    res.second = index[(1 << m) - 1];
  }
  for (int x = 0; x < (1 << m); x++) {
    if (index[x] == -1) continue;
    for (int y = 0; y < (1 << m); y++) {
      if (index[y] == -1) continue;
      if ((x | y) == (1 << m) - 1) {
        res.first = index[x] + 1;
        res.second = index[y] + 1;
        return res;
      }
    }
  }
  return res;
}
void Solve() {
  cin >> n >> m;
  a.resize(n, vector<int>(m, 0));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> a[i][j];
    }
  }
  int left = 0;
  int right = (int)1e9;
  pair<int, int> ans = {-1, -1};
  while (left <= right) {
    int mid = left + (right - left) / 2;
    pair<int, int> res = Check(mid);
    if (res != make_pair(-1, -1)) {
      left = mid + 1;
      ans = res;
    } else {
      right = mid - 1;
    }
  }
  cout << ans.first << ' ' << ans.second << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  Solve();
  return 0;
}