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].
#include <bits/stdc++.h>

using namespace std;
int n,q,k;
int* a;
int* p;
int main()
{
    cin>>n>>q>>k;
    a=new int[n];
    p=new int[n];
    for(int i=0;i<n;i++){
        cin>>a[i];
        p[i]=0;
    }
    if(n==1){
        for(int i=0;i<q;i++){
            int l,r;
            cin>>l>>r;
            cout<<k-1<<endl;
        }
    }
    else{
        p[0]=a[1]-2;
        for(int i=1;i<n-1;i++){
            p[i]=a[i+1]-a[i-1]-2;
        }
        p[n-1]=k-a[n-2]-1;
        int ps[n];
        ps[0]=p[0];
        for(int i=1;i<n;i++)
            ps[i]=p[i]+ps[i-1];

      /*  for(int i=0;i<n;i++){
            cout<<p[i]<<' ';
        }
        cout<<endl;
        for(int i=0;i<n;i++)
            cout<<ps[i]<<' ';
        cout<<endl;*/
        for(int i=0;i<q;i++){
            int l,r;
            cin>>l>>r;
            l--,r--;
            if(l==r){
                cout<<k-1<<endl;
            }
            else{
                int res=ps[r-1]+k-a[r-1]-1-ps[l]+a[l+1]-2;
                cout<<res<<endl;
            }
        }

    }
    return 0;
}
