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].
// Starting with the name of Allah

#include <bits/stdc++.h>
#define ll long long
#define nl endl
#define Nayon ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define gcd(a,b)         __gcd(a,b)
#define lcm(a,b)         (a/gcd(a,b))*b
#define v_sum(a)         accumulate(a.begin(),a.end(),0)
#define v_max(a)         *max_element(a.begin(),a.end())
#define v_min(a)         *min_element(a.begin(),a.end())
#define v_uni(a)         a.erase(unique(a.begin(),a.end()),a.end())
#define pi          	 3.14159265358979323846
#define bit(n)			  __builtin_popcount (n)
using namespace std;

ll const mod=1e9+7;



int main()
{
	Nayon;
	ll t,n,m,i,j,k,cnt=0,sum1=0,ccnt=0,x,y,z=1,x1,y1=0,z1,total=0,sum=0,x2,y2,sz,sum2=0;
	cin>>n>>t>>k;
	ll a[n+2];
	for(i=1;i<=n;i++){
		cin>>a[i];
	}
	while(t--){
		cin>>x>>y;
		cout<<k+(a[y]-a[x]+1)-2*(y-x+1)<<nl;
	}
	return 0;
}

//Happy Coding............
