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.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.io.StreamTokenizer;


public class A {
    private static StreamTokenizer in;
    private static PrintWriter out;
    
    private static int nextInt() throws Exception{
        in.nextToken();
        return (int)in.nval;
    }
    
    private static String nextString() throws Exception{
        in.nextToken();
        return in.sval;
    }
    
    static{
        in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        out = new PrintWriter(System.out);
    }
    
    public static void main(String[] args) throws Exception{
        int n = nextInt();
        int[] mas = new int[n];
        int[] m = new int[n];
        for(int i = 0; i<n; i++){
            mas[i] = nextInt();
        }
        for(int i = 0; i<n - 1; i++){
            m[i] = Math.abs(mas[i] - mas[i+1]);
            
        }
        m[n - 1] = Math.abs(mas[n -1] - mas[0]);
        int max = 2000;
        int maxi = 0;
        for(int i = 0; i<n - 1; i++){
            if(m[i] < max){max = m[i];
            maxi = i+1;}
        }
        if(m[n - 1]<max){
            maxi = n;
            System.out.println((n)+" "+1);
            return;
        }
        System.out.println(maxi+" "+ (maxi+1));
    }
}
