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>
#define ll long long
#define pb push_back
#define pf push_front
#define s second
#define f first
#define MOD 1000000007
#define MOD2 998244353
#define MAX 100000
using namespace std;
vector<int> gr[200005];
bool visited[200005];
//int lev[200005],rem[200005];
/*void dfs(int node,int par,int level)
{
        lev[node]=level;
        int c=0;
        for(auto it:gr[node])
        {
                if(it==par)
                        continue;
                dfs(it,node,level+1);
                c+=rem[it];
        }
        rem[node]=1+ c;
}*/
void dfs(int node)
{
     visited[node]=true;
     for(auto it:gr[node])
     {
             if(!visited[it])
                dfs(it);
     }
}
int main()
{
ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
ll n,q,k;
cin>>n>>q>>k;
ll *a=new ll[n+1];
for(ll i=1;i<n+1;i++)
        cin>>a[i];
ll st[n+1],en[n+1],mid[n+1];
for(ll i=1;i<n;i++)
{
        st[i]=a[i+1]-2;
}
for(ll i=2;i<=n;i++)
{
        en[i]=k-a[i-1] -1;
}
mid[1]=0;
mid[n]=0;
for(ll i=2;i<n;i++)
{
        mid[i]=mid[i-1] + a[i+1]-a[i-1]-2;
}
while(q--)
{
        ll l,r;
        cin>>l>>r;
        if(l==r)
                cout<<k-1<<endl;
        else
        {
             cout<<st[l]+en[r] + (mid[r-1]-mid[l])<<endl;
        }
}
return 0;}