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.*;

final public class B {

    public static void main(String[] args) throws Exception{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringBuilder sb = new StringBuilder();
        String[] str = br.readLine().split(" ");
        int n = Integer.parseInt(str[0]), q = Integer.parseInt(str[1]), k = Integer.parseInt(str[2]);
        StringTokenizer st = new StringTokenizer(br.readLine());
        int[] arr = new int[n+2];
        for(int i = 1; i<=n; i++){
            arr[i] = Integer.parseInt(st.nextToken());
        }
        arr[n+1] = k;
        int[] max = new int[n+1];
        max[1] = arr[2]-1;
        for(int i = 2; i+1<=n; i++){
            max[i] = arr[i+1] - arr[i-1] - 2 + max[i-1];
        }
        max[n] = max[n-1] + (k-arr[n-1]-1);
        for(int i = 0; i<q; i++){
            String[] qr = br.readLine().split(" ");
            int l = Integer.parseInt(qr[0]), r = Integer.parseInt(qr[1]);
            long ans = 0;
            if(r == l){
                sb.append(k-1).append('\n');
                continue;
            }
            else if(r-l == 1){
                ans = arr[r] - 2 + (k-arr[l]-1);
            }
            else {
                ans = arr[l+1]-2 + (k - arr[r-1] -1);
                ans += max[r-1] - max[l];
            }
            sb.append(ans).append('\n');
        }
        System.out.println(sb);
    }
}
