Given a positive integer k, two arrays are called k-similar if:

  * they are strictly increasing; 
  * they have the same length; 
  * all their elements are positive integers between 1 and k (inclusive); 
  * they differ in exactly one position. 



You are given an integer k, a strictly increasing array a and q queries. For each query, you are given two integers l_i ≤ r_i. Your task is to find how many arrays b exist, such that b is k-similar to array [a_{l_i},a_{l_i+1}…,a_{r_i}]. 

Input

The first line contains three integers n, q and k (1≤ n, q ≤ 10^5, n≤ k ≤ 10^9) — the length of array a, the number of queries and number k.

The second line contains n integers a_1, a_2, …,a_n (1 ≤ a_i ≤ k). This array is strictly increasing — a_1 < a_2 < … < a_n.

Each of the following q lines contains two integers l_i, r_i (1 ≤ l_i ≤ r_i ≤ n).

Output

Print q lines. The i-th of them should contain the answer to the i-th query.

Examples

Input


4 2 5
1 2 4 5
2 3
3 4


Output


4
3


Input


6 5 10
2 4 6 7 8 9
1 4
1 2
3 5
1 6
5 5


Output


8
9
7
6
9

Note

In the first example:

In the first query there are 4 arrays that are 5-similar to [2,4]: [1,4],[3,4],[2,3],[2,5].

In the second query there are 3 arrays that are 5-similar to [4,5]: [1,5],[2,5],[3,5].

import java.util.Scanner;

public class Main {


    public static void main(String[] args) {

        int n,q,k;
        int a[] = new int[100010];
        int d[] = new int[100010];
        long s[] = new long[100010];
        Scanner in = new Scanner(System.in);
        n = in.nextInt();
        q = in.nextInt();
        k = in.nextInt();
        for (int i = 1; i <= n; i ++) a[i] = in.nextInt();
        for (int i = 2; i < n; i ++) d[i] = a[i+1]-a[i-1];
        s[2] = d[2];
        for (int i = 3; i < n; i ++) s[i] = d[i] + s[i-1];
        for (int i = 1; i <= q; i ++) {
            int l = in.nextInt();
            int r = in.nextInt();
            if (l == r) {
                System.out.println(k-1);
                continue;
            }
            long ans = a[l+1]-2;
            ans += s[r-1] - s[l] - 2*(r-l-1);
            ans += k-a[r-1]-1;
            System.out.println(ans);
        }

    }




}
 				 		 	 					  		 			  	  		