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;



int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    long long n,q,k;
    cin>>n>>q>>k;
    deque<long long>a(n);
    for(int i=0;i<n;i++) cin>>a[i];
    a.push_front(0);

    deque<long long>start(n,0);
    start[0]=a[2]-1-1;
    for(int i=1;i<n-1;i++) start[i]=a[i+2]-1-a[i]-1;
    start[n-1]=k-a[n-1]-1;
    start.push_front(0);
    for(int i=1;i<n+1;i++) start[i]+=start[i-1];
   // for(auto i:start) cout<<i<<" ";cout<<endl;
    while(q--)
    {
        long long l,r;cin>>l>>r;
        if(l==r){cout<<k-1<<"\n";continue;}
        long long ans=start[r-1]-start[l];
        ans+=k-a[r-1]-1;
        ans+=a[l+1]-1-1;
        cout<<ans<<"\n";
    }
}