#include <bits/stdc++.h>
#define int long long
using namespace std;

signed main(){
    ios_base::sync_with_stdio(false);cin.tie(NULL);
    int n,q,k;
    cin >> n >> q >> k;
    vector <int> A(n+2);
    A[0]=0;
    A[n+1]=k+1;
    for(int i=1;i<=n;i++)cin >> A[i];
    vector <int> P(n+2);
    for(int i=1;i<=n;i++){
        P[i]=(A[i]-A[i-1]-1)+(A[i+1]-A[i]-1);
    }
    for(int i=1;i<=n;i++){
        P[i]+=P[i-1];
    }
    // for(int i=0;i<n+2;i++)cout << P[i] << " ";
    for(int i=0;i<q;i++){
        int l,r;
        cin >> l >> r;
        int ans=P[r]-P[l-1]+(k+1-A[r+1])+A[l-1];
        cout << ans << '\n';
    }
}
### Prompt

Your challenge is to write a CPP solution to the following problem:

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].

### Response

```cpp
#include <bits/stdc++.h>
#define int long long
using namespace std;

signed main(){
    ios_base::sync_with_stdio(false);cin.tie(NULL);
    int n,q,k;
    cin >> n >> q >> k;
    vector <int> A(n+2);
    A[0]=0;
    A[n+1]=k+1;
    for(int i=1;i<=n;i++)cin >> A[i];
    vector <int> P(n+2);
    for(int i=1;i<=n;i++){
        P[i]=(A[i]-A[i-1]-1)+(A[i+1]-A[i]-1);
    }
    for(int i=1;i<=n;i++){
        P[i]+=P[i-1];
    }
    // for(int i=0;i<n+2;i++)cout << P[i] << " ";
    for(int i=0;i<q;i++){
        int l,r;
        cin >> l >> r;
        int ans=P[r]-P[l-1]+(k+1-A[r+1])+A[l-1];
        cout << ans << '\n';
    }
}
```