#include <bits/stdc++.h>

using namespace std;


int main()
{
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long int n,q,k,i;
  cin>>n>>q>>k;
  long int arr[n],diff[n];
  for(i=0;i<n;i++)
  cin>>arr[i];
  diff[0]=0;
  for(i=1;i<n;i++)
  {
      diff[i]=diff[i-1]+2*(arr[i]-arr[i-1]-1);
  }
  while(q--)
  {
      long int l,r;
      cin>>l>>r;
      l--;
      r--;
      long int ans=0;
      if(arr[l]-1>0)
      ans+=arr[l]-1;
      if(k-arr[r]>0)
      ans+=k-arr[r];
      ans+=diff[r]-diff[l];
      cout<<ans<<endl;
  }
}

### Prompt

In cpp, your task is to solve 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>

using namespace std;


int main()
{
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long int n,q,k,i;
  cin>>n>>q>>k;
  long int arr[n],diff[n];
  for(i=0;i<n;i++)
  cin>>arr[i];
  diff[0]=0;
  for(i=1;i<n;i++)
  {
      diff[i]=diff[i-1]+2*(arr[i]-arr[i-1]-1);
  }
  while(q--)
  {
      long int l,r;
      cin>>l>>r;
      l--;
      r--;
      long int ans=0;
      if(arr[l]-1>0)
      ans+=arr[l]-1;
      if(k-arr[r]>0)
      ans+=k-arr[r];
      ans+=diff[r]-diff[l];
      cout<<ans<<endl;
  }
}

```