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 <class T>
void read(T &u) {
  u = 0;
  char c = getchar();
  int flag = 0;
  while (c < '0' || c > '9') flag |= (c == '-'), c = getchar();
  while (c >= '0' && c <= '9')
    u = (u << 3) + (u << 1) + (c ^ 48), c = getchar();
  if (flag) u = -u;
}
template <class T>
void write(T u) {
  if (u < 0) printf("-"), u = -u;
  if (u < 10)
    putchar(48 + u);
  else
    write(u / 10), putchar(48 + u % 10);
}
void write(char u) { putchar(u); }
template <class T, class... Arg>
void read(T &u, Arg &...v) {
  read(u);
  read(v...);
}
template <class T, class... Arg>
void write(T u, Arg... v) {
  write(u);
  write(v...);
}
template <class T>
bool checkmin(T &a, T b) {
  return (b < a ? a = b, 1 : 0);
}
template <class T>
bool checkmax(T &a, T b) {
  return (b > a ? a = b, 1 : 0);
}
template <class T>
T _min(T a, T b) {
  return (a < b ? a : b);
}
template <class T>
T _max(T a, T b) {
  return (a > b ? a : b);
}
const int mod = 1e9 + 7, N = 300005;
int n, m, mx = 0, full;
int a[N][8];
int vis[300];
void init() {
  read(n, m);
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < m; ++j) {
      read(a[i][j]);
      checkmax(mx, a[i][j]);
    }
  }
  full = (1 << m) - 1;
}
pair<int, int> check(int x);
void solve() {
  int l = 0, r = mx, mid, best = -1;
  while (l <= r) {
    mid = (l + r) >> 1;
    if (check(mid).first) {
      best = mid;
      l = mid + 1;
    } else {
      r = mid - 1;
    }
  }
  pair<int, int> ans = check(best);
  write(ans.first, ' ', ans.second, '\n');
}
int main() {
  int t = 1;
  while (t--) {
    init();
    solve();
  }
  return 0;
}
pair<int, int> check(int x) {
  for (int i = 0; i <= full; ++i) {
    vis[i] = 0;
  }
  for (int i = 1; i <= n; ++i) {
    int tmp = 0;
    for (int j = 0; j < m; ++j) {
      if (a[i][j] < x) continue;
      tmp |= (1 << j);
    }
    vis[tmp] = i;
  }
  for (int i = 0; i <= full; ++i) {
    for (int j = 0; j <= full; ++j) {
      if ((i | j) == full && vis[i] && vis[j]) {
        return make_pair(vis[i], vis[j]);
      }
    }
  }
  return make_pair(0, 0);
}