n soldiers stand in a circle. For each soldier his height ai is known. A reconnaissance unit can be made of such two neighbouring soldiers, whose heights difference is minimal, i.e. |ai - aj| is minimal. So each of them will be less noticeable with the other. Output any pair of soldiers that can form a reconnaissance unit.

Input

The first line contains integer n (2 ≤ n ≤ 100) — amount of soldiers. Then follow the heights of the soldiers in their order in the circle — n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 1000). The soldier heights are given in clockwise or counterclockwise direction.

Output

Output two integers — indexes of neighbouring soldiers, who should form a reconnaissance unit. If there are many optimum solutions, output any of them. Remember, that the soldiers stand in a circle.

Examples

Input

5
10 12 13 15 10


Output

5 1


Input

4
10 20 30 40


Output

1 2
import java.util.*;

public class E {

    public static void main(String[] args) {

        Scanner fk = new Scanner(System.in);

        int n = fk.nextInt();
        int first = fk.nextInt();
        int a = first;
        int min_index =0;
        int min = Integer.MAX_VALUE;

        int[] arr = new int[n];

        for(int i =1; i<n; i++) {
            int t = fk.nextInt();
            arr[i-1] = Math.abs(a-t);
            a = t;
        }
        arr[n-1] = Math.abs(first-a);

        for(int i =0; i<n; i++){
            if(arr[i] < min){
                min = arr[i];
                min_index = i+1;
            }
        }

       if(min_index+1 == n)
           System.out.println(min_index +" "+ ++min_index);
       else
           System.out.println(min_index+" "+(min_index+1)%n);


      /*  String s = fk.next();
        int len = s.length();
        Stack<Character> stack = new Stack<>();
        int max = 0;
        int counter = 0;

        for(int i =0; i<len; i++){
            if(s.charAt(i) == '(') {
                stack.push('(');
            }else if(stack.size() == 0) {
                max = Math.max(max, counter);
                counter = 0;
            }
            else {
                stack.pop();
                counter++;
            }
        }

        max = Math.max(max, counter);

        System.out.println(2*max);
*/
    }

}

