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 + 2;
const int M = 1e6 + 2;
bool vis[N];
int in[M];
void solve() {
  int n;
  scanf("%d", &n);
  long long ret = 1e18;
  int x = -1, y = -1;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &in[i]);
    if (vis[in[i]]) {
      if (ret > in[i]) {
        ret = in[i];
        x = y = in[i];
      }
    }
    vis[in[i]] = true;
  }
  int lim = (int)1e7;
  int num, j, k;
  for (num = 1; num <= lim && ret > num; ++num) {
    int tx = -1, ty = -1;
    for (j = num; j <= lim && ty == -1; j += num) {
      if (vis[j]) {
        if (tx == -1) {
          tx = j;
        } else {
          ty = j;
        }
      }
    }
    if (ty != -1) {
      long long g = (long long)tx * ty / num;
      if (ret > g) {
        ret = g;
        x = tx, y = ty;
      }
    }
  }
  int rx = -1, ry;
  for (int i = 0; i < n; ++i) {
    if (in[i] == x && rx == -1) {
      rx = i + 1;
    } else if (in[i] == y) {
      ry = i + 1;
    }
  }
  if (rx > ry) {
    swap(rx, ry);
  }
  printf("%d %d\n", rx, ry);
}
int main() { solve(); }
