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].
using namespace std;
#include<bits/stdc++.h>
#define ll long long
#define Case int t ; cin>>t ; while(t--)
#define all(v) ((v).begin()),((v).end())
#define sz(n) n.size()
#define fx(x) fixed<<setprecision(x)
#define pir pair<int,int>
ll mod = 1000000007; ;
int dx[] = {-1, 1, 0, 0, -1, 1, -1, 1};
int dy[] = {0, 0, 1, -1, 1,  -1, -1,1};
void fast (){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
}
ll gcd(ll x, ll y){
    if(!y) return x;
    return gcd(y, x%y);
}
ll lcm(ll x, ll y){
    return (x/gcd(x, y))*y ;
}
ll fastpow(ll a, ll b){
    ll res = 1 ;
    while(b)
    {
        if(b%2) res*=a ;
        b/=2, a*=a ;
    }
    return res ;
}
bool isprime (ll n){
    if(n==1 || n==0) return 0 ;
    for(int i = 2 ; i<=sqrt(n) ; i++) if(n%i==0) return 0;
    return 1 ;
}
void solve(){
    ll ans = 0 , n , q , k ; cin>>n>>q>>k ;
    vector<ll> num(n+1) , pre(n+1) , arr(n+2) ;
    for(int i = 1 ; i<=n ; i++)cin >> num[i] ;
    num.push_back(num[n]+1) ;
    for(int i = 1 ; i<=n ; i++){
        if(i==n) arr[i] = (num[i] - num[i-1] - 1) + (k - num[i]) ;
        else arr[i] = (num[i] - num[i-1] - 1) + (num[i+1] - num[i] - 1) ;
        pre[i] = arr[i]+pre[i-1] ;
    }
    while(q--){
        int l , r ; cin>>l>>r ;
        if(l == r) {cout << k-1 << "\n" ;  continue ;}
        ans = (pre[r] - pre[l-1]) - (arr[l]+arr[r])+(num[l]-1 + (num[l+1] - num[l] - 1))+((num[r]-num[r-1]-1)+(k-num[r])) ;
        cout << ans << "\n" ;
    }
}
int main(){
    fast() ;
    solve();
}
