You are given an array a consisting of n integers a_1, a_2, ..., a_n.

Your problem is to find such pair of indices i, j (1 ≤ i < j ≤ n) that lcm(a_i, a_j) is minimum possible.

lcm(x, y) is the least common multiple of x and y (minimum positive number such that both x and y are divisors of this number).

Input

The first line of the input contains one integer n (2 ≤ n ≤ 10^6) — the number of elements in a.

The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^7), where a_i is the i-th element of a.

Output

Print two integers i and j (1 ≤ i < j ≤ n) such that the value of lcm(a_i, a_j) is minimum among all valid pairs i, j. If there are multiple answers, you can print any.

Examples

Input


5
2 4 8 3 6


Output


1 2


Input


5
5 2 11 3 7


Output


2 4


Input


6
2 5 10 1 10 2


Output


1 4
#include <bits/stdc++.h>
using namespace std;
const int lmt = 1e7 + 7;
bool vis[lmt];
long long L = 1e18, idx = -1, idy = -1;
int id[lmt], n;
long long a[lmt];
void sieve() {
  for (long long i = 1; i < lmt; i++) {
    long long f = -1, s = -1;
    for (int j = i; j < lmt; j += i) {
      if (vis[j]) {
        if (f == -1)
          f = j;
        else if (s == -1)
          s = j;
      }
      if (f == -1 || s == -1) continue;
      break;
    }
    if (f == -1 || s == -1) continue;
    long long now = (f * s) / i;
    if (L > now) {
      L = now;
      idx = id[f], idy = id[s];
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (id[a[i]] > 0 && L > a[i]) {
      L = a[i], idx = id[a[i]], idy = i;
    }
    id[a[i]] = i;
    vis[a[i]] = 1;
  }
  sieve();
  cout << min(idx, idy) << " " << max(idx, idy) << "\n";
  return 0;
}
