# include<iostream>
# include<algorithm>

using namespace std;

int n, q, k;

const int N = 1e5+10;

int a[N];

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    
    cin>>n>>q>>k;
    for(int i=1; i<=n; ++i)   cin>>a[i];
    
    while(q--)
    {
        int l, r;
        int ans = 0;
        cin>>l>>r;
        
        if(l==r)
        {
            cout<<k-1<<'\n';
            continue;
        }
        
        ans += (a[l+1] - 1 - 1);
        ans += (k - a[r-1] - 1);
        ans += (a[r] + a[r-1] - a[l] - a[l+1] - (r-l-1)*2);

        
        cout<<ans<<'\n';
    }
    
    return 0;
}
### Prompt

Create a solution in CPP for 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>
# include<algorithm>

using namespace std;

int n, q, k;

const int N = 1e5+10;

int a[N];

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    
    cin>>n>>q>>k;
    for(int i=1; i<=n; ++i)   cin>>a[i];
    
    while(q--)
    {
        int l, r;
        int ans = 0;
        cin>>l>>r;
        
        if(l==r)
        {
            cout<<k-1<<'\n';
            continue;
        }
        
        ans += (a[l+1] - 1 - 1);
        ans += (k - a[r-1] - 1);
        ans += (a[r] + a[r-1] - a[l] - a[l+1] - (r-l-1)*2);

        
        cout<<ans<<'\n';
    }
    
    return 0;
}
```