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.*;
public class codeforces1{
 
 public static void main(String[] args)
 {
 	 Scanner sc=new Scanner(System.in);
 	 
        int n=sc.nextInt();
      	 
      	  int q=sc.nextInt();
         
        int k=sc.nextInt();
        int i,j;
        int arr[]=new int[n+1];
        long newarr[]=new long[n+1];
        
        
        for(i=1;i<=n;i++)
         arr[i]=sc.nextInt();
         
         arr[0]=0;

       long sum=0l,p=0l;
       newarr[0]=0;
        for(i=1;i<=n;i++)
        {
         
            if(i==1&&n!=1)
            {
             p=arr[i+1]-1-1;
             sum+=p;
             newarr[i]=sum;
            } 
            
            else if(i==n)
            {
               p=k-arr[i-1]-1;
                sum+=p;
                newarr[i]=sum;
            
            }
            
            else
            {
               p=arr[i+1]-arr[i-1]-2;
                sum+=p;
                newarr[i]=sum;
             }
        
        }
       
       for(i=1;i<=q;i++)
       {
       
         int l=sc.nextInt();
         int r=sc.nextInt();
         
           if(l==r)
            System.out.println(k-1);
         
         else
         {
         long ans=arr[l+1]-1-1+k-arr[r-1]-1;
          long s=newarr[r-1]-newarr[l];
         
         System.out.println(ans+s);
         }
      
       }   
        
     
     
     }
     
     
   }  
