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 {

    public static void main (String[] args) throws Exception {

        String s = "6 5 10\n" +
                "2 4 6 7 8 9\n" +
                "1 4\n" +
                "1 2\n" +
                "3 5\n" +
                "1 6\n" +
                "5 5";

//        String s = "6 1 10\n" +
//                "2 4 6 7 8 9\n" +
//                "1 4";

//        String s = "1\n" +
//                "-3 -5\n" +
//                "LDLDLDDDR\n";

//        br = new BufferedReader(new StringReader(s));
        br = new BufferedReader(new InputStreamReader(System.in));
        bw = new BufferedWriter(new OutputStreamWriter(System.out));

        rl(); int n=nin(), q=nin(), k=nin();
        rl(); int a[] = new int[n];
        for (int i=0; i<n; ++i) a[i]=nin();
        long S[] = new long[n];
        long sum=0;
        int p=a[0]+1;
        for (int j=1; j<n-1; ++j) {
            int c=0;
            sum+=(a[j]-p);
            int nxt=a[j+1]-1;
            sum+=(nxt-a[j]);
            p=a[j]+1;
            S[j] = sum;
        }

        for (int i=0; i<q; ++i) {
            rl(); int l=nin()-1, r=nin()-1;

            long c=0;
            p=1;
            c+=(a[l]-p);
            int nxt;
            if (l==n-1 || l==r) {
                nxt=k;
            } else {
                nxt=a[l+1]-1;
            }
            c+=(nxt-a[l]);

            if (l<r) {
                p=a[r-1]+1;
                c+=(a[r]-p);
                nxt=k;
                c+=(nxt-a[r]);
            }

            if (r-l>0) {
                c+=(S[r-1]-S[l]);
            }

            bw.write(c+"\n");
        }


        bw.flush();
    }

    static BufferedReader br; static BufferedWriter bw; static StringTokenizer st;
    static void rl() throws Exception{ st = new StringTokenizer(br.readLine()); }
    static long nlo(){ return Long.parseLong(st.nextToken()); }
    static int nin(){ return Integer.parseInt(st.nextToken()); }
    /*private static void te(){}*/
    static double ndo(){ return Double.parseDouble(st.nextToken()); }
    static char[] nca(){ return st.nextToken().toCharArray(); }
}

