#include<iostream>

using namespace std;

#define ll long long

int main()
{
    int n,q,k; cin >> n >> q >> k;
    int a[n];
    for (int i = 0; i < n; i++) cin >> a[i];
    int s[n];
    s[0] = 0;
    for (int i = 1; i < n; i++)
    {
        s[i]=s[i-1]+a[i]-a[i-1]-1;
    }
    for (int i = 0; i < q; i++)
    {
        int l,r; cin >> l >> r;
        cout << 2*(s[r-1]-s[l-1])+a[l-1]+k-a[r-1]-1 << endl;
    }
    
}
### Prompt

Generate 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<iostream>

using namespace std;

#define ll long long

int main()
{
    int n,q,k; cin >> n >> q >> k;
    int a[n];
    for (int i = 0; i < n; i++) cin >> a[i];
    int s[n];
    s[0] = 0;
    for (int i = 1; i < n; i++)
    {
        s[i]=s[i-1]+a[i]-a[i-1]-1;
    }
    for (int i = 0; i < q; i++)
    {
        int l,r; cin >> l >> r;
        cout << 2*(s[r-1]-s[l-1])+a[l-1]+k-a[r-1]-1 << endl;
    }
    
}
```