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>
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
const int N = 10000005;
int pos[N];
inline long long min(const long long &a, const long long &b) {
  return a < b ? a : b;
}
inline void chmin(long long &a, const long long &b) { (a > b) && (a = b); }
int main() {
  int n, x, p1 = 0, p2 = 0;
  long long ans = 0x7fffffffffffffff;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &x);
    if (pos[x]) {
      if (ans > x) {
        ans = x;
        p1 = pos[x], p2 = i;
      }
    } else {
      pos[x] = i;
    }
  }
  for (register int i = 1; i <= 1e7; ++i) {
    register int tmp = 0, p = 0;
    if (i >= ans) break;
    for (register int j = i; j <= 1e7; j += i) {
      if (pos[j]) {
        if (tmp) {
          if (ans > 1ll * j * tmp / i) {
            ans = 1ll * j * tmp / i;
            p1 = pos[j], p2 = p;
          }
        } else {
          tmp = j;
          p = pos[j];
        }
      }
    }
  }
  if (p1 < p2)
    printf("%d %d\n", p1, p2);
  else
    printf("%d %d\n", p2, p1);
  return 0;
}
