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;
int a[1000005];
vector<int> v[10000001];
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  int i1 = 0, i2 = 0, mx = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    mx = max(mx, a[i]);
    v[a[i]].push_back(i);
  }
  long long ans = (long long)(1e18 + 7);
  int s1 = 0, s2 = 0;
  for (int i = 1; i <= mx; i++) {
    vector<int> xy;
    if (v[i].size()) xy.push_back(i);
    if (v[i].size() > 1) xy.push_back(i);
    for (int j = i + i; j <= mx; j = j + i) {
      if (xy.size() == 2) break;
      if (v[j].size()) {
        if (xy.size() == 1) {
          xy.push_back(j);
        } else {
          if (v[j].size() > 1)
            xy.push_back(j), xy.push_back(j);
          else
            xy.push_back(j);
        }
      }
    }
    if (xy.size() != 2)
      continue;
    else {
      long long xx = ((1ll * xy[0] * xy[1]) / i);
      if (ans > xx) ans = xx, s1 = xy[0], s2 = xy[1];
    }
  }
  i1 = v[s1][0];
  if (s1 == s2)
    i2 = v[s2][1];
  else
    i2 = v[s2][0];
  cout << min(i1, i2) << ' ' << max(i1, i2);
}
