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.*;
import java.util.*;

public class Solution {

    StreamTokenizer in;
    PrintWriter out;

    public static void main(String[] args) throws Exception {
        new Solution().run();
    }

    public void run () throws Exception {
        in = new StreamTokenizer (new BufferedReader(new InputStreamReader(System.in)));
        out = new PrintWriter(new OutputStreamWriter(System.out));
        solve();
        out.flush();
    }

    int nextInt() throws Exception {
        in.nextToken();
        return (int) in.nval;
    }

    public void solve() throws Exception {
        int n = nextInt();
        int [] a = new int [n];
        for (int i=0; i<n;i++)
        {
            a[i]=nextInt();
        }
        int min=Integer.MAX_VALUE;
        int pos1=0;
        int pos=0;
        for (int i=0; i<n-1;i++)
        {
            if (Math.abs(a[i]-a[i+1])<min)
            {
                min = Math.abs(a[i] - a[i + 1]);
                pos=i;
                pos1=i+1;
            }
        }
        if (Math.abs(a[n-1]-a[0])<min)
        {
            min = Math.abs(a[n - 1] - a[0]);
            pos = n-1;
            pos1=0;
        }
        out.println((pos+1)+" "+(pos1+1));
    }

}
