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;
long long gcd(long long n, long long m) {
  if (n == 0) return m;
  return gcd(m % n, n);
}
bool ifprime(long long n) {
  if (n == 1) return false;
  long long i;
  for (i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
void disp(vector<long long> v) {
  for (auto u : v) cout << u << " ";
  cout << "\n";
}
void remleadzero(vector<int> &v) {
  vector<int> u;
  int i = 0;
  while (v[i] == 0) i++;
  while (i < v.size()) {
    u.push_back(v[i]);
    i++;
  }
  v = u;
}
long long fast_expo(long long n, long long m, long long md) {
  long long a = 1;
  while (m > 0) {
    if (m & 1) a = (a * n) % md;
    n = (n * n) % md;
    m /= 2;
  }
  return a % md;
}
long long sqroot(long long n) {
  long double N = n;
  N = sqrtl(N);
  long long sq = N - 2;
  sq = max(sq, 0LL);
  while (sq * sq < n) {
    sq++;
  }
  if ((sq * sq) == n) return sq;
  return -1;
}
long long a[300005][8], cur_val[300005];
long long n, m;
bool poss(long long k) {
  long long val = (1 << m) - 1;
  set<long long> val_set;
  for (int i = 0; i < n; i++) {
    long long cur = 0;
    for (int j = 0; j < m; j++) {
      if (a[i][j] >= k) cur |= (1 << j);
    }
    val_set.insert(cur);
    cur_val[i] = cur;
  }
  bool found = 0;
  for (long long i = 0; i < n; i++) {
    for (auto u : val_set) {
      if ((cur_val[i] | u) == (val)) {
        found = true;
        break;
      }
    }
    if (found) break;
  }
  return found;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  ;
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> a[i][j];
    }
  }
  long long lo = 0, hi = 1e9, mid;
  while (lo <= hi) {
    mid = (lo + hi) >> 1;
    if (poss(mid))
      lo = mid + 1;
    else
      hi = mid - 1;
  }
  long long val = (1 << m) - 1;
  set<long long> val_set;
  for (int i = 0; i < n; i++) {
    long long cur = 0;
    for (int j = 0; j < m; j++) {
      if (a[i][j] >= hi) cur |= (1 << j);
    }
    val_set.insert(cur);
    cur_val[i] = cur;
  }
  bool found = 0;
  long long ans1 = -1, ans2 = -1, val1 = 0, val2 = 0;
  for (long long i = 0; i < n; i++) {
    for (auto u : val_set) {
      if ((cur_val[i] | u) == (val)) {
        ans1 = i;
        val1 = u;
        found = true;
        break;
      }
    }
    if (found) break;
  }
  cout << ans1 + 1 << ' ';
  for (int i = 0; i < n; i++) {
    if (cur_val[i] == val1) {
      cout << i + 1;
      return 0;
    }
  }
  return 0;
}