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 <cstdio>
#include <bits/stdc++.h>
#include "iostream"
#include <algorithm>
using namespace std;
#define int long long
#define INF (int)10e9+10
#define MAX_INT ((int)10e5+5)

void solve(){
    int n,q,k;
    cin>>n>>q>>k;
    vector<int> a(n);
    for (int i = 0; i < n; ++i) {
        cin>>a[i];
    }
    vector<int> possibility(n),first(n),last(n);
    for (int i = 0; i < n; ++i) {
        if(i!=0 || i!=n-1){
            possibility[i] = (a[i+1]-a[i-1]-2);
        }
        if(i<n-1)
            first[i] = a[i+1]-2;
        if(i>0)
            last[i] = k - a[i-1] -1;
    }
    vector<int> pre(n+1,0);
    for (int i = 1; i <= n-1; ++i) {
        pre[i] = pre[i-1] + possibility[i];
    }
    pre[n]=pre[n-1];
    while(q--){
        int l1,r1;
        cin>>l1>>r1;
        if(l1==r1){
            //cout<<"loop 1 ";
            cout<<k-1<<endl;
        }
        else if(l1+1==r1){
            //cout<<"loop 2 ";
            cout<<first[l1-1]+last[r1-1]<<endl;
        }
        else{
            //cout<<"loop 3 ";
            cout<<first[l1-1]+pre[r1-2]-pre[l1-1]+last[r1-1]<<endl;
        }
    }

}


int32_t main() {
    ios::sync_with_stdio(false);
    int t=1;
    //cin>>t;
    while(t--){
        solve();
    }
    return 0;
}