You are given an array a consisting of n integer numbers.

Let instability of the array be the following value: max_{i = 1}^{n} a_i - min_{i = 1}^{n} a_i.

You have to remove exactly one element from this array to minimize instability of the resulting (n-1)-elements array. Your task is to calculate the minimum possible instability.

Input

The first line of the input contains one integer n (2 ≤ n ≤ 10^5) — the number of elements in the array a.

The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5) — elements of the array a.

Output

Print one integer — the minimum possible instability of the array if you have to remove exactly one element from the array a.

Examples

Input


4
1 3 3 7


Output


2


Input


2
1 100000


Output


0

Note

In the first example you can remove 7 then instability of the remaining array will be 3 - 1 = 2.

In the second example you can remove either 1 or 100000 then instability of the remaining array will be 100000 - 100000 = 0 and 1 - 1 = 0 correspondingly.
import java.util.HashMap;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int arr[] = new int[sc.nextInt()];
        int min=Integer.MAX_VALUE,min2=Integer.MAX_VALUE,max1=Integer.MIN_VALUE,max2=Integer.MIN_VALUE;
        for(int i =0 ;i< arr.length;i++)
        {
            arr[i] = sc.nextInt();
            if(arr[i]<min2){
                if(arr[i]<min){
                    min2 = min;
                    min = arr[i];

                }
                else
                {
                    min2 = arr[i];
                }
            }
            if(arr[i]>max2){
                if(arr[i]>max1){
                    max2 = max1;
                    max1 = arr[i];

                }
                else{
                    max2 = arr[i];
                }
            }
        }
       if(min2-min<max1-max2){
           System.out.println(max2-min);
       }
       else{
           System.out.println(max1-min2);
       }
    }
}
