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.io.*;
import java.util.*;

public class cfrcs {
// 
//  static boolean prime[] = new boolean[n+1]; 
//	static void sieve(int n) 
//    { 
//        
//        for(int i=0;i<n;i++) 
//            prime[i] = true; 
//          
//        for(int p = 2; p*p <=n; p++) 
//        { 
//            
//            if(prime[p] == true) 
//            { 
//       
//                for(int i = p*p; i <= n; i += p) 
//                    prime[i] = false; 
//            } 
//        } 
//    } 

	static int cn(int a, int b) {
		int count = 0;
		while (a != 0) {
			a = a / b;
			count++;
		}
		return count;
	}

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		sc = new Scanner(System.in);
		int n=sc.nextInt();
		int q=sc.nextInt();
		int k=sc.nextInt();
		int arr[]=new int[n];
		for(int i=0;i<n;i++) {
			arr[i]=sc.nextInt();
		}
		long b[]=new long[n];
		for(int i=0;i<n-1;i++) {
			if(i==0) {
				int aa=arr[i+1]-1-1;
				b[i]=aa;
			}
			else {
				int aa=arr[i+1]-arr[i-1]-1-1;
				b[i]=aa;
			}
		}
		if(n!=1) {
			b[n-1]=k-arr[n-1-1]-1;
		}
		else if(n==1) {
			b[n-1]=k-1;
		}
		long c[]=new long[n];
		c[0]=b[0];
		for(int i=1;i<n;i++) {
			c[i]+=c[i-1]+b[i];
		}
//		for(int i=0;i<n;i++) {
//			System.out.print(b[i]+" ");
//		}System.out.println();
//		for(int i=0;i<n;i++) {
//			System.out.print(c[i]+" ");
//		}System.out.println();
		
		
		while(q-->0) {
			int l=sc.nextInt();
			int r=sc.nextInt();
			l--;r--;
			if(l==r) {
				System.out.println(k-1);
			}
			else if(n==1)System.out.println(k-1);
			else{long ans=c[r-1]-c[l];
			long aa=arr[l+1]-1-1;
			long bb=k-arr[r-1]-1;
			ans=ans+aa+bb;
			System.out.println(ans);
			}
			
		}
		

	}

}
