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.*;
import java.util.Comparator;
import java.util.*;

public class Main {


    public static void main(String[] args) {
        Scanner scanner =  new Scanner(System.in);
        int n = scanner.nextInt();
        int q = scanner.nextInt();
        int k = scanner.nextInt();
        int[] a = new int[n];
        int[] update = new int[n];
        for(int i=0;i<n;i++)a[i]=scanner.nextInt();
        if(n>1){update[0] = a[1]-2; update[n-1] = k-a[n-2]-1;}
        else update[0] = k-1;
        for(int i =1;i<n-1;i++)update[i] = a[i+1]-2-a[i-1];
        int[] prev = new int[n+1];
        for(int i = 1;i<=n;i++){
            prev[i] = prev[i-1]+update[i-1];
        }

        while(q-->0){
            int l = scanner.nextInt()-1;
            int r = scanner.nextInt()-1;
            if(l==r){System.out.println((k-1));continue;}
            int ans = 0;
            ans = prev[r]-prev[l+1];
            //System.out.println("CHECK 1 "+ans);
            ans+=k-a[r-1]-1;
            //System.out.println("CHECK 2 "+ans);
            ans+=a[l+1]-2;
            //System.out.println("CHECK 3 "+ans);
            System.out.println(ans);
        }



    }
}
