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[]) {
        Scanner s = new Scanner(System.in);
        int n = s.nextInt();
        int q = s.nextInt();
        int k = s.nextInt();
        int arr[] = new int[n + 1];
        for(int i = 1; i <= n; i++){
            arr[i] = s.nextInt();
        }
        int dp[] = new int[n + 1];
        for(int i = 1; i <= n; i++){
            dp[i] = dp[i - 1] + 2 * (arr[i] - arr[i - 1] - 1);

        }
        while(q-- != 0){
            int l = s.nextInt();
            int r = s.nextInt();
            System.out.println(arr[l] - 1 + k - arr[r] + dp[r] - dp[l]);
        }
    }
}