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>
std::pair<int, int> DR[] = {{-1, 0}, {0, 1},   {1, 0}, {0, -1},
                            {-1, 1}, {-1, -1}, {1, 1}, {1, -1}};
using namespace std;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  ~debug() { cerr << endl; }
  template <class c>
  typename enable_if<sizeof dud<c>(0) != 1, debug&>::type operator<<(c i) {
    cerr << boolalpha << i;
    return *this;
  }
  template <class c>
  typename enable_if<sizeof dud<c>(0) == 1, debug&>::type operator<<(c i) {
    return *this << range(begin(i), end(i));
  }
  template <class c, class b>
  debug& operator<<(pair<b, c> d) {
    return *this << "(" << d.first << ", " << d.second << ")";
  }
  template <class c>
  debug& operator<<(rge<c> d) {
    *this << "[";
    for (auto it = d.b; it != d.e; ++it) *this << ", " + 2 * (it == d.b) << *it;
    return *this << "]";
  }
};
int gcd(int a, int b) {
  if (b) return gcd(b, a % b);
  return a;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
pair<int, int> val[1 << 19];
int n, m, a[1 << 19][10];
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cerr.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> a[i][j];
    }
  }
  int fullmask = (1 << m) - 1;
  for (int mask = 0; mask < (1 << m); mask++) {
    vector<int> pos;
    for (int j = 0; j < m; j++) {
      if (mask & (1 << j)) pos.push_back(j);
    }
    pair<int, int> mx;
    for (int i = 0; i < n; i++) {
      int mn = 1e9;
      for (auto it : pos) {
        mn = min(mn, a[i][it]);
      }
      if (!i) mx = make_pair(mn, i);
      mx = max(mx, make_pair(mn, i));
    }
    val[mask] = mx;
  }
  pair<int, pair<int, int>> ans = make_pair(-1, make_pair(0, 0));
  for (int mask = 0; mask <= fullmask; mask++) {
    ans = max(ans, make_pair(min(val[mask].first, val[fullmask - mask].first),
                             make_pair(val[mask].second,
                                       val[fullmask - mask].second)));
  }
  cout << ans.second.first + 1 << ' ' << ans.second.second + 1 << '\n';
}