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;
namespace chhokmah {
int a[10000007], ps1[10000007], ps2[10000007];
int n, as1, as2;
long long ans = (1ll << 60);
void solve(int MAXN) {
  for (int i = 1; i <= MAXN; i++) {
    if (i >= ans) break;
    int s1 = 0, s2 = 0;
    for (int j = i; j <= MAXN; j += i) {
      if (!a[j]) continue;
      if (!s1)
        s1 = j, s2 = a[j];
      else {
        long long lcm = 1ll * s1 * j / i;
        if (ans > lcm) {
          ans = lcm;
          as1 = s2;
          as2 = a[j];
          break;
        }
      }
    }
  }
}
void chhokmah() {
  memset(a, 0, sizeof(a));
  scanf("%d", &n);
  for (int i = 1, x; i <= n; i++) {
    scanf("%d", &x);
    if (a[x] && x * 1ll <= ans)
      ans = x, as1 = a[x], as2 = i;
    else
      a[x] = i;
  }
  solve(10000007 - 1);
  if (as1 > as2) swap(as1, as2);
  cout << as1 << " " << as2 << endl;
}
}  // namespace chhokmah
int main() {
  chhokmah::chhokmah();
  return 0;
}
