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 Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << ": " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << ": " << arg1 << " |";
  __f(comma + 1, args...);
}
pair<long long, long long> b[(1 << 9)];
long long c1, c2;
long long n, m;
long long a[300005][10];
bool check(long long v) {
  long long te;
  c1 = -1;
  c2 = -1;
  for (long long i = 0; i < n; i++) {
    te = 0;
    for (long long j = 0; j < m; j++)
      if (a[i][j] >= v) te += (1 << j);
    te = (~te) & ((1 << m) - 1);
    if (b[te].first >= v) {
      c1 = i;
      c2 = b[te].second;
    }
  }
  if (c1 == -1) {
    return 0;
  } else {
    return 1;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> m;
  memset(b, 0, sizeof(b));
  for (long long i = 0; i < n; i++)
    for (long long j = 0; j < m; j++) cin >> a[i][j];
  long long s = 0, e = 100000000000, mi, te;
  for (long long i = 0; i < (1 << m); i++) {
    b[i] = {0, 0};
    for (long long j = 0; j < n; j++) {
      te = INT_MAX;
      for (long long k = 0; k < m; k++) {
        if ((1 << k) & i) {
          te = min(te, a[j][k]);
        }
      }
      if (te > b[i].first && te != INT_MAX) {
        b[i] = {te, j};
      }
    }
  }
  while (s < e) {
    mi = (s + e) / 2;
    if (check(mi)) {
      s = mi;
    } else {
      e = mi;
    }
    if (s == e - 1) {
      if (check(e)) {
        s = e;
      } else {
        e = s;
      }
    }
  }
  check(s);
  long long ma = 0, in1 = 0;
  for (long long i = 0; i < n; i++) {
    te = INT_MAX;
    for (long long j = 0; j < m; j++) te = min(te, a[i][j]);
    if (te > ma) {
      in1 = i + 1;
      ma = te;
    }
  }
  if (ma > s) {
    cout << in1 << " " << in1 << "\n";
  } else {
    cout << c1 + 1 << " " << c2 + 1 << "\n";
  }
  return 0;
}