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 p[10000005];
int a[1000005];
int n;
int main() {
  scanf("%d", &n);
  int ma = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    p[a[i]]++;
    ma = max(ma, a[i]);
  }
  int x, y, cnt = 0;
  long long ans = (long long)1e18;
  int sx = 0, sy = 0;
  for (int i = 1; i <= ma; i++) {
    cnt = x = y = 0;
    for (int j = i; j <= ma; j += i) {
      if (p[j] >= 2 && cnt == 0) {
        x = y = j;
        break;
      } else if (p[j] && cnt == 1) {
        cnt++;
        y = j;
        break;
      } else if (p[j] && cnt == 0) {
        cnt++;
        x = j;
      }
    }
    if (x > 0 && y > 0) {
      long long tmp = 1ll * x * y / i;
      if (ans > tmp) {
        ans = tmp;
        sx = x;
        sy = y;
      }
    }
  }
  int xx = 0, yy = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] == sx) {
      xx = i;
      break;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (a[i] == sy && xx != i) {
      yy = i;
      break;
    }
  }
  if (xx > yy) swap(xx, yy);
  printf("%d %d\n", xx, yy);
  return 0;
}
