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 long long maxn = 1e7 + 10;
long cnt[maxn], id[maxn];
long n;
long fir, sec;
long long res;
long fw;
int main() {
  scanf("%ld", &n);
  fir = sec = -1;
  res = maxn * maxn;
  long i, j;
  for (i = 1; i <= n; ++i) {
    scanf("%ld", &j);
    fw = max(j, fw);
    if (cnt[j] && j < res) {
      res = (long long)j;
      fir = id[j];
      sec = i;
    }
    cnt[j] = 1;
    id[j] = i;
  }
  long nfir, nsec;
  long long nres = 0ll;
  for (i = 1; i <= fw; ++i) {
    nfir = nsec = 0;
    nres = (long long)maxn * maxn;
    for (j = i; j <= fw; j += i) {
      if (cnt[j] && nfir) {
        nsec = j;
        break;
      } else if (cnt[j])
        nfir = j;
    }
    nres = (long long)nfir / i * nsec;
    if (!nres) continue;
    if (res > nres) {
      res = nres;
      fir = id[nfir];
      sec = id[nsec];
    }
  }
  if (fir > sec) swap(fir, sec);
  printf("%ld %ld\n", fir, sec);
  return 0;
}
