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;
static const int maxn = 1e7 + 5;
map<long long int, vector<int> > Map;
long long int freq[maxn], arr[maxn];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  long long int maxi = 0;
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
    freq[arr[i]]++;
    Map[arr[i]].push_back(i);
    maxi = max(maxi, arr[i]);
  }
  long long int lcm = 1e18;
  int p = -1, q = -1;
  for (int i = 1; i <= maxi; i++) {
    int cnt = 0;
    int pos1 = -1;
    int pos2 = -1;
    for (int j = i; j <= maxi && cnt < 2; j += i) {
      if (freq[j] == 0) continue;
      if (cnt == 0) {
        {
          cnt = 1;
          pos1 = Map[j][0];
        }
        if (Map[j].size() > 1) {
          cnt = 2;
          pos2 = Map[j][1];
        }
      } else {
        pos2 = Map[j][0];
        cnt = 2;
      }
    }
    if (cnt < 2) continue;
    assert(pos1 > 0 && pos2 > 0);
    long long int new_lcm = arr[pos1] * arr[pos2] / i;
    if (new_lcm < lcm) lcm = new_lcm, p = pos1, q = pos2;
  }
  cout << min(p, q) << " " << max(p, q) << '\n';
}
