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 maxn = 1e7 + 10;
int f[maxn], s[maxn], a[maxn];
vector<int> V;
int main() {
  int n, Max = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (f[a[i]] == 0)
      f[a[i]] = i;
    else
      s[a[i]] = i;
    Max = max(Max, a[i]);
  }
  long long ans = (1ll) << 60;
  int p1, p2;
  for (int i = 1; i <= Max; i++) {
    V.clear();
    for (int j = i; j <= Max; j += i) {
      if (f[j]) V.push_back(f[j]);
      if (s[j]) V.push_back(s[j]);
      if (V.size() >= 2) break;
    }
    if (V.size() < 2) continue;
    if (ans > 1ll * a[V[0]] * a[V[1]] / i)
      ans = 1ll * a[V[0]] * a[V[1]] / i, p1 = V[0], p2 = V[1];
  }
  if (p1 > p2) swap(p1, p2);
  cout << p1 << " " << p2 << endl;
  return 0;
}
