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 maxn = 1e7;
int a[maxn + 10];
long long lcm = 1e14;
int pos[maxn + 10], in1, in2;
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (!pos[a[i]])
      pos[a[i]] = i;
    else {
      if (lcm > 1LL * a[i]) {
        lcm = 1LL * a[i];
        in1 = i;
        in2 = pos[a[i]];
      }
    }
  }
  int first, item;
  for (int i = 1; i <= maxn; i++) {
    if (i > lcm) break;
    first = item = 0;
    for (int j = i; j <= maxn; j += i) {
      if (!pos[j]) continue;
      if (!first) {
        first = j;
        item = pos[j];
      } else {
        if (lcm > 1LL * first * j / i) {
          lcm = 1LL * first * j / i;
          in1 = item;
          in2 = pos[j];
        }
        break;
      }
    }
  }
  if (in1 > in2) in1 ^= in2 ^= in1 ^= in2;
  printf("%d %d\n", in1, in2);
}
