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.StringTokenizer;

public class B_ReplaceAndKeepSorted {
    private static final BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    private static final PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
    private static StringTokenizer st;

    private static int readInt() throws IOException {
        while (st == null || !st.hasMoreTokens()) st = new StringTokenizer(br.readLine());
        return Integer.parseInt(st.nextToken());
    }

    public static void main(String[] args) throws IOException {
        int n = readInt();
        int q = readInt();
        int k = readInt();
        int[] a = new int[n + 1];
        for (int i = 1; i <= n; i++) {
            a[i] = readInt();
        }

        for (int i = 0; i < q; i++) {
            int l = readInt();
            int r = readInt();
            int ans = k + 1 - 2 * (r - l + 1) - a[l] + a[r];
            pw.println(ans);
        }
        pw.close();
    }
}

/*
a1, a2, ..., an
a1-1 + (a2-a1-1)
        (a2-1-1) + (a3-a2-1)
                   (a3-a2-1) + (a4-a3-1)
                                      + (an-an_1-1)
                                      + (an-an_1-1) + (k - an)
ans +(a1-1)+ (k-an) = 2 * ( k - n)
ans = 2*(k-n) - k - a1 +1 +an = k+1 - 2 * n - a1 + an
 */