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 double eps = 1e-6;
const int dir[8][2] = {-2, -1, -2, 1, 2, -1, 2, 1, -1, 2, -1, -2, 1, 2, 1, -2};
const int maxn = 1e7 + 5;
int cnt[maxn];
int w[1000000 + 5];
int n;
int maxx = -1;
int main() {
  cin >> n;
  memset(cnt, 0, sizeof(cnt));
  for (int i = 1; i <= n; i++) {
    scanf("%d", &w[i]);
    cnt[w[i]]++;
    maxx = max(maxx, w[i]);
  }
  long long ans1 = -1, ans2 = -1, ans = 214700000000483647;
  for (int i = 1; i <= maxx; i++) {
    long long a = -1, b = -1;
    for (int j = i; j <= maxx; j += i) {
      if (cnt[j] >= 2) {
        if (a == -1) a = j;
        if (b == -1) b = j;
        if (a / i * b < ans) {
          ans = a / i * b;
          ans1 = a;
          ans2 = b;
        }
        break;
      }
      if (cnt[j]) {
        if (a == -1)
          a = j;
        else if (b == -1) {
          b = j;
          if (a / i * b < ans) {
            ans = a / i * b;
            ans1 = a;
            ans2 = b;
          }
          break;
        }
      }
    }
  }
  int a, b;
  for (int i = 1; i <= n; i++)
    if (w[i] == ans1) {
      a = i;
      w[i] = -1;
      break;
    }
  for (int i = 1; i <= n; i++)
    if (w[i] == ans2) {
      b = i;
      break;
    }
  cout << min(a, b) << " " << max(a, b) << endl;
  return 0;
}
