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;
inline long long gcd(long long a, long long b) {
  while (b != 0) {
    long long c = a % b;
    a = b;
    b = c;
  }
  return a < 0 ? -a : a;
}
inline long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
inline long long lowbit(long long x) { return x & (-x); }
const double PI = 3.14159265358979323846;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const long long mod = 998244353;
inline char nc() {
  static char buf[1000000], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1000000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline long long rd() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
const double eps = 1e-8;
const int M = 4e6 + 10;
const int N = 3e5 + 10;
int a[N][10];
int n, m;
int f[1 << 9];
int m1, m2;
int maxx;
bool check(int x) {
  memset(f, 0, sizeof f);
  for (int i = 1; i <= n; i++) {
    int res = 1;
    int sum = 0;
    for (int j = 0; j < m; j++) {
      if (a[i][j] >= x) sum += res;
      res <<= 1;
    }
    f[sum] = i;
  }
  for (int i = 1; i <= maxx; i++) {
    if (!f[i]) continue;
    for (int j = i; j <= maxx; j++) {
      if (!f[j]) continue;
      if ((i | j) == maxx) {
        m1 = f[i];
        m2 = f[j];
        return 1;
      }
    }
  }
  return 0;
}
int main() {
  n = rd(), m = rd();
  m1 = m2 = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < m; j++) {
      a[i][j] = rd();
    }
  }
  maxx = (1 << m) - 1;
  int l = 0, r = 1e9;
  while (l < r) {
    int mid = (l + r + 1) >> 1;
    if (check(mid))
      l = mid;
    else
      r = mid - 1;
  }
  printf("%d %d\n", m1, m2);
}