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 N = 1e7 + 10;
const int M = 1e9 + 7;
const int inf = 1e9 + 7;
const long long INF = 1e18;
int a[N];
int n;
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  int x;
  long long ans = INF;
  int lx, rx;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    if (a[x] && ans > x) {
      ans = x;
      lx = a[x];
      rx = i;
    }
    a[x] = i;
  }
  for (int i = 1; i < N; i++) {
    if (i >= ans) break;
    int l = -1, r;
    for (int j = i; j < N; j += i) {
      if (!a[j]) continue;
      if (l == -1) {
        l = j;
      } else {
        r = j;
        long long tmp = l / i;
        tmp *= r;
        if (tmp < ans) {
          ans = tmp;
          lx = a[l];
          rx = a[r];
        }
      }
    }
  }
  printf("%d %d\n", min(lx, rx), max(lx, rx));
}
