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<iostream>
#include<stdio.h>
#include<bits/stdc++.h>
using namespace std;
#define endl '\n'

int main(){

    ios_base::sync_with_stdio(false);cin.tie(NULL);
   
    #ifndef ONLINE_JUDGE 
    freopen("input.txt", "r", stdin); 
    freopen("output.txt", "w", stdout); 
    #endif 
   
    int n,q,k;
    cin>>n>>q>>k;
    int arr[n];
    for(int i=0;i<n;i++)
        cin>>arr[i];

    int dp[n];
    for(int i=0;i<n-1;i++){
        if(i>0){
            int a=arr[i]-arr[i-1];
            int b=arr[i+1]-arr[i];
            a--,b--;
            dp[i]=a+b;
        }
        else
        {
            int a=arr[i+1]-arr[i];
            a--;
            dp[i]=a;
        }
    }
    if(n>1)
        dp[n-1]=arr[n-1]-arr[n-2]-1;
      
    int pre=0;
    for(int i=1;i<n;i++){
        pre=dp[i-1];
        dp[i]+=pre;
    }

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

            else
                ans+=dp[b-1]-dp[a-1]+(arr[a]-arr[a-1]-1)+(arr[a-1]-1)+k-arr[b-1];

            cout<<ans<<endl;
        }

    }
}