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;
int mp[10000007];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  int a[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    mp[a[i]]++;
  }
  long long int res = 1e16, N = 1e7 + 1;
  int ans[2];
  for (int gc = 1; gc < min(res, N); gc++) {
    long long int gf = 0, gs = 0;
    for (int j = gc; j < min(res, N); j += gc) {
      if (mp[j] && gf) {
        gs = j;
        break;
      } else if (mp[j]) {
        gf = j;
        if (mp[j] > 1) {
          gs = j;
          break;
        }
      }
    }
    if (((gf * gs) / gc) < res && gf != 0 && gs != 0) {
      res = (gf * gs) / gc;
      ans[0] = gf;
      ans[1] = gs;
    }
  }
  int fl1 = -1, fl2 = -1;
  for (int i = 0; i < n; i++) {
    if (a[i] == ans[0] && fl1 == -1) {
      fl1 = i + 1;
    } else if (a[i] == ans[1] && fl2 == -1) {
      fl2 = i + 1;
    }
  }
  cout << min(fl1, fl2) << " " << max(fl1, fl2);
}
