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 int N = 1e7 + 5;
long long int n, a[1000050], mx, ans = 1e17, x1, x2, in1, in2;
vector<long long int> d[N];
pair<long long int, long long int> solve() {
  cin >> n;
  for (long long int i = 1; i <= n; i++) {
    cin >> a[i];
    mx = max(mx, a[i]);
  }
  for (long long int i = 1; i <= n; i++) {
    if (d[a[i]].size() > 1) continue;
    d[a[i]].push_back(a[i]);
  }
  for (long long int i = 1; i <= mx; i++) {
    for (long long int j = i + i; j <= mx; j += i) {
      if (d[i].size() > 1) break;
      for (long long int k : d[j]) {
        if (d[i].size() > 1) break;
        d[i].push_back(k);
      }
    }
  }
  for (long long int i = 1; i <= mx; i++) {
    if (d[i].size() <= 1) continue;
    long long int temp = (d[i][0] * d[i][1]) / i;
    if (temp < ans) {
      ans = temp;
      x1 = d[i][0];
      x2 = d[i][1];
    }
  }
  for (long long int i = 1; i <= n; i++) {
    if (a[i] == x1 && (!in1))
      in1 = i;
    else if (a[i] == x2)
      in2 = i;
  }
  if (in1 > in2) swap(in1, in2);
  return {in1, in2};
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  pair<long long int, long long int> P = solve();
  cout << P.first << " " << P.second;
  return 0;
}
