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
import java.util.Arrays;
import java.util.Scanner;

public class Main { // G
	/**
	 * lcm = a*b/gcd, thus we can enumerate gcd
	 * @param args
	 */
	
	static int[] cnt = new int[10000000+1];
	
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int[] a = new int[n];
		int m = 0;
		for(int i=0; i<n; ++i) {
			a[i] = in.nextInt();
			++cnt[a[i]];
			m = Math.max(m, a[i]);
		}
		long lcm = Long.MAX_VALUE;
		int[] candidate = null;
		for(int gcd=1; gcd<=m; ++gcd) {
			int[] num = new int[2];
			boolean isFirst = true;
			for(int i=gcd; i<=m; i+=gcd) {
				if(cnt[i]==0) {
					continue;
				}
				if(!isFirst) {
					num[1] = i;
					break;
				} else {
					num[0] = i;
					isFirst = false;
					if(cnt[i]>1) {
						num[1] = i;
						break;
					}
				}
			}
			if(num[1] > 0) {
				long x = ((long)num[0])*num[1]/gcd;
				if(x<lcm) {
					lcm = x;
					candidate = num;
				}
			}
		}
		int[] pos = new int[2];
		for(int i=0;i<2;++i) {
			for(int j=0;j<n;++j) {
				if(a[j]==candidate[i]) {
					a[j] = -1;
					pos[i] = j+1;
					break;
				}
			}
		}
		Arrays.sort(pos);
		System.out.println(pos[0] + " " + pos[1]);
	}
	
}
