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;
typedef long long ll;
typedef long long int lli;
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define rep(i,a,b) for(int i=a ; i<b; i++)
#define T ll t;cin>>t;while(t--)
#define Pi 3.141592653589793238
#define MOD 1000000007
#define vii vector<pair<int,int>>
#define inf 1e8
#define FAST ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
const int MAX = (int)1e5 + 5;
int a[MAX];
void solve()
{
    int n,q,k,l,r;
    int sum=0,res;
    cin>>n>>q>>k;
    rep(i,0,n)
     cin>>a[i];
     rep(i,0,q)
     {
         cin>>l>>r;
         r--; l--;
         cout<<(a[r]-a[l]+1)+k-2*((r+1)-(l+1)+1)<<"\n";
     }
}
int main()
{

	 solve();
	cout << "\n";
	return 0;
}
