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;
#ifndef ONLINE_JUDGE
    #include "debug.h"
#endif
#define int long long 
const int mod = 1e9 + 7, mxn = 2e5 + 1, inf = LLONG_MAX;

int solve(){
    int n, k, q;
    cin>> n>> q>> k;
    vector<int> a(n);
    for(int i = 0; i < n; i++){
        cin>> a[i];
    }
    vector<int> cnt(n, 0);
    for(int i = 1; i < n - 1; i++){
        cnt[i] = (a[i] - a[i - 1] - 1) + (a[i + 1] - a[i] - 1);
    }
    for(int i = 1; i < n; i++){
        cnt[i] += cnt[i - 1];
    }
    while(q--){
        int l, r;
        cin>> l>> r;
        if(l == r){
            cout<< (k - 1)<< "\n";
            continue;
        }
        l--, r--;
        int ans = a[l] - 1 + a[l + 1] - a[l] - 1 + k - a[r] + a[r] - a[r - 1] - 1 + cnt[r - 1] - cnt[l];
        cout<< ans<< "\n"; 
    }
    return 0;
}

signed main(){
    ios_base::sync_with_stdio(NULL);cin.tie(NULL);cout.tie(NULL);
    int t = 1;
    // cin>> t;
    while(t--){
        solve();
    }
}