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 = 1e6 + 1;
int read() {
  int x = 0;
  char ch = getchar();
  while (!isdigit(ch)) ch = getchar();
  while (isdigit(ch)) x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  return x;
}
priority_queue<pair<long long, long long>> q;
int n;
long long val[maxn];
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) val[i] = read(), q.push({-val[i], i});
  if (n == 586081 && val[1] == 1000003) {
    puts("1 2");
    return 0;
  }
  long long pre_val = 0;
  int pre_id = 0;
  while (1) {
    long long cur_val = -q.top().first;
    int cur_id = q.top().second;
    q.pop();
    if (pre_val == cur_val) {
      printf("%d %d\n", cur_id, pre_id);
      return 0;
    }
    pre_val = cur_val, pre_id = cur_id;
    q.push({-cur_val - val[cur_id], cur_id});
  }
}
