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 long long l_inf = numeric_limits<long long>::max();
const long long l_ninf = numeric_limits<long long>::min();
const int inf = numeric_limits<int>::max();
const int ninf = numeric_limits<int>::min();
const double pi = 1.00 * acos(-1.00);
const double eps = 1e-8;
void cio() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * (b / gcd(a, b))); }
int sq(int x) { return x * x; }
const int maxn = 3e5 + 5;
int n, m;
int a[maxn][10];
pair<int, int> check(int mid) {
  pair<int, int> ret = {-1, -1};
  set<int> st;
  map<int, int> id;
  for (int i = 0; i < n; i++) {
    int x = 0;
    for (int j = 0; j < m; j++) {
      if (a[i][j] >= mid) x |= (1 << j);
    }
    if (!st.count(x)) {
      st.insert(x);
      id[x] = i;
    }
  }
  for (auto it1 = st.begin(); it1 != st.end(); it1++) {
    for (auto it2 = st.begin(); it2 != st.end(); it2++) {
      int x = *it1, y = *it2;
      if ((x | y) == (1 << m) - 1) {
        ret.first = id[x], ret.second = id[y];
        break;
      }
    }
  }
  return ret;
}
int main() {
  scanf("%d%d", &n, &m);
  int r = 0;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) {
      scanf("%d", &a[i][j]);
      r = max(r, a[i][j]);
    }
  int l = 0;
  pair<int, int> ans;
  while (l <= r) {
    int mid = (l + r) >> 1;
    auto c = check(mid);
    if (c.first != -1) {
      ans = c;
      l = mid + 1;
    } else
      r = mid - 1;
  }
  printf("%d %d\n", ans.first + 1, ans.second + 1);
  ;
}