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;
const int INF = 1061109567;
const long long LINF = 4557430888798830399ll;
const int MOD = 1000000007;
template <typename T>
void read(T &x) {
  x = 0;
  char c = getchar();
  if (c == EOF) exit(0);
  while (!isdigit(c)) c = getchar();
  while (isdigit(c)) {
    x = x * 10 + (c - '0');
    c = getchar();
  }
  assert(c != EOF);
}
int reads(char *s) {
  int l = 0;
  *s = getchar();
  if (*s == EOF) exit(0);
  while (*s != ' ' && *s != '\t' && *s != '\r' && *s != '\n' && *s != EOF) {
    ++s;
    ++l;
    *s = getchar();
  }
  assert(*s != EOF);
  *s = '\0';
  return l;
}
long long qpow(long long x, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) res = res * x % MOD;
    x = x * x % MOD;
    n >>= 1;
  }
  return res;
}
int a[300005][8];
int vis[1 << 8];
int b[2400005], btop;
int ansi, ansj;
int n, m;
bool check_help(int l, int r, int mid) {
  for (int i = l; i <= r; ++i) {
    int cur = 0;
    for (int j = 0; j < m; ++j) cur = (cur << 1) | (a[i][j] >= mid);
    vis[cur] = i;
  }
  int all1 = (1 << m) - 1;
  for (int i = all1; i; --i) {
    if (!vis[i]) continue;
    for (int j = 0; j < m; ++j) {
      if ((i >> j) & 1) vis[i ^ (1 << j)] = vis[i];
    }
  }
  for (int i = 0; i < (1 << m); ++i) {
    if (!vis[i]) continue;
    int w = all1 ^ i;
    if (vis[w]) {
      ansi = vis[i];
      ansj = vis[w];
      return true;
    }
  }
  return false;
}
bool check(int mid) {
  memset(vis, 0, sizeof(vis));
  int m1 = n / 4;
  int m2 = n / 2;
  int m3 = n * 3 / 4;
  if (check_help(1, m1, mid) || check_help(m1 + 1, m2, mid) ||
      check_help(m2 + 1, m3, mid) || check_help(m3 + 1, n, mid))
    return true;
  return false;
}
void bs() {
  int L = 1, R = btop;
  while (true) {
    int mid = (L + R) >> 1;
    if (mid == L) {
      if (check(R))
        return;
      else {
        assert(check(L));
        return;
      }
    }
    if (check(mid))
      L = mid;
    else
      R = mid - 1;
  }
}
void test_case() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < m; ++j) {
      read(a[i][j]);
      b[++btop] = a[i][j];
    }
  }
  sort(b + 1, b + 1 + btop);
  btop = unique(b + 1, b + 1 + btop) - (b + 1);
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < m; ++j)
      a[i][j] = lower_bound(b + 1, b + 1 + btop, a[i][j]) - b;
  }
  bs();
  printf("%d %d\n", ansi, ansj);
}
int main() {
  int t = 1;
  for (int i = 1; i <= t; ++i) {
    test_case();
  }
}