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].
                               /* When you get down to nothing.. */
#include<bits/stdc++.h>
#define sync            ios_base::sync_with_stdio(0);cin.tie(NULL);
#define ll              long long
#define ull             unsigned ll
#define ld              long double
#define vi              vector<int>
#define vll             vector<ll> 
#define vvi             vector<vector<int>>
#define vvll            vector<vector<ll>>
#define all(x)          x.begin(),x.end()
#define FOR(i,a,b)      for(int i=a;i<b;++i)
using namespace std;
 
const int MAX=1e6+5;
const int mod=1e9+7;
const ld pi=acos(-1);

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());


int main(){
    // #ifndef ONLINE_JUDGE 
    //     freopen("input.txt", "r", stdin);
    //     freopen("output.txt", "w", stdout);
    // #endif
    clock_t clk = clock();
    sync 
    // int t;
    // cin>>t;
    // while(t--){
        //test cases starts here
        ll n,q,k;
        cin>>n>>q>>k;
        vll a(n+1);
        FOR(i,1,n+1)  cin>>a[i];
        while(q--){
            ll l,r;
            cin>>l>>r;
            ll ans=0;
            if(l == r){
                ans=k-1;
                cout<<ans<<endl;
                continue;
            }
            ans+=(a[l+1]-2)+(k-a[r-1]-1);
            if(r - l > 1){
                ll t=a[r-1]+a[r]-a[l]-a[l+1]-2*(r-l-1);
                ans+=t;
            }
            cout<<ans<<endl;
        }

    // }

    // cerr << '\n'<<"Time (in s): " << double(clock() - clk) * 1.0 / CLOCKS_PER_SEC << '\n';
}

                                /* God is always upto something! */
