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;
typedef long long ll;
// bool isprime(ll n) 
// { 
//     // Corner case 
//     if (n <= 1) 
//         return false; 
  
//     // Check from 2 to n-1 
//     for (ll i = 2; i <= n/2; i++) 
//         if (n % i == 0) 
//             return false; 
  
//     return true; 
// } 
//  bool cmp(pair<int, int>& a, 
//          pair<int, int>& b) 
// {
//     return a.second < b.second; 
// }
// void query(ll x, ll n , ll ar[])
// {
//     if(x>=1 && x<=n)
//     cout<<"? "<<x<<endl;;
//     cout.flush();
//     cin>>ar[x];
// }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n,q,k,i;
  cin>>n>>q>>k;
  int arr[n];
  int brr[n];
  for(i=0;i<n;i++)
  cin>>arr[i];
  int pr[n];
   for(i=0;i<n;i++)
      { if(i==0)
      brr[i]=arr[i+1]-2;
      else if(i==n-1)
      brr[i]=k-arr[i-1]-1;
      else
      brr[i]=arr[i+1]-arr[i-1]-2;
      } 
      pr[0]=brr[0];
     // cout<<pr[0]<<" ";
  for(i=1;i<n;i++){
     pr[i]=pr[i-1]+brr[i];
    // cout<<pr[i]<<" ";
  }
//  cout<<endl;

    while(q--)
  {
      int a,b;
      cin>>a>>b;
      a--;
      b--;
      ll ans=0;
      if(a==b)
    {  cout<<k-1<<endl;
    continue;
    }
    if(a>0)
ans=pr[b]-pr[a-1];
else
ans=pr[b];
if(b!=n-1)
ans+=k-arr[b+1]+1;
if(a!=0)
ans+=arr[a-1];
      
      cout<<ans<<endl;
  
  }
  
        return 0;
    }
    