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 MAXN 100005
using namespace std;

int n,q;
long long k,a[MAXN];
long long sum[MAXN];

template<typename Drake>
void Read(Drake &x){
	x=0;char ch=getchar();
	while(!isdigit(ch)){
		ch=getchar();
	}
	while(isdigit(ch)){
		x=(x<<1)+(x<<3)+(ch^48);
		ch=getchar();
	}
}
 
int main(){
	Read(n);Read(q);Read(k);
	for(int i=1;i<=n;++i)Read(a[i]);
	sum[0]=0;
	sum[1]=a[2]-2;
	for(int i=2;i<n;++i)
	sum[i]=sum[i-1]+a[i+1]-a[i-1]-2;
	sum[n]=sum[n-1]+k-a[n-1]-1;
	while(q--){
		int l,r;long long ans;
		Read(l);Read(r);
		if(l==r){
			printf("%lld\n",k-1);
			continue;
		} 
		ans=sum[r-1]-sum[l];
		ans+=a[l+1]-2;
		ans+=k-a[r-1]-1;
		printf("%lld\n",ans);
	} 
	return 0;
}
