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>
#pragma GCC optimize("-O3")
#pragma GCC optimize("-Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long long N1 = 1e6 + 5, N2 = 1e7 + 5;
long long n, a[N1], cur1 = -1, cur2 = -1, ind1 = -1, ind2 = -1, ans = LLONG_MAX,
                    ans1, ans2, mx;
vector<long long> v[N2];
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  for (long long i = 1; i <= n; i++)
    cin >> a[i], v[a[i]].push_back(i), mx = max(mx, a[i]);
  for (long long i = 1; i <= mx; i++) {
    cur1 = cur2 = ind1 = ind2 = -1;
    for (long long j = i; j <= mx; j += i)
      if (v[j].size()) {
        if (cur1 == -1) {
          cur1 = j, ind1 = v[j][0];
          if (v[j].size() > 1) {
            cur2 = j, ind2 = v[j][1];
            long long cur_lcm = (cur1 / i) * cur2;
            if (cur_lcm < ans) ans = cur_lcm, ans1 = ind1, ans2 = ind2;
            cur1 = cur2 = ind1 = ind2 = -1;
            break;
          }
        } else {
          cur2 = j, ind2 = v[j][0];
          long long cur_lcm = (cur1 / i) * cur2;
          if (cur_lcm < ans) ans = cur_lcm, ans1 = ind1, ans2 = ind2;
          cur1 = cur2 = ind1 = ind2 = -1;
          break;
        }
      }
  }
  if (ans1 > ans2) swap(ans1, ans2);
  cout << ans1 << " " << ans2;
  return 0;
}
