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 = 1e6 + 10;
const int MAXX = 1e7 + 10;
int cnt[MAXX];
int A[MAXN];
int main() {
  int n;
  scanf("%d", &n);
  int mx = -1;
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    A[i] = x;
    cnt[x]++;
    mx = max(mx, x);
  }
  long long ans = 1e18;
  for (int i = 1; i <= mx; i++) {
    long long n1 = -1, n2 = -1;
    for (int j = i; j <= mx && (n1 == -1 || n2 == -1); j += i) {
      for (int k = 0; k < min(2, cnt[j]) && (n1 == -1 || n2 == -1); k++) {
        if (n1 == -1)
          n1 = j;
        else if (n2 == -1)
          n2 = j;
      }
    }
    if (n2 != -1) {
      ans = min(ans, n1 / i * n2);
    }
  }
  int rem = 2;
  for (int i = 1; i <= n && rem > 0; i++) {
    if (ans % A[i] == 0) {
      printf("%d ", i);
      rem--;
    }
  }
  return 0;
}
