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<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define rg register
inline int read(){
	rg int x=0,fh=1;
	rg char ch=getchar();
	while(ch<'0' || ch>'9'){
		if(ch=='-') fh=-1;
		ch=getchar();
	}
	while(ch>='0' && ch<='9'){
		x=(x<<1)+(x<<3)+(ch^48);
		ch=getchar();
	}
	return x*fh;
}
const int maxn=1e5+5;
int n,q,k,a[maxn];
long long sum[maxn];
int main(){
	n=read(),q=read(),k=read();
	for(rg int i=1;i<=n;i++) a[i]=read();
	a[n+1]=k+1;
	for(rg int i=1;i<=n;i++){
		sum[i]=sum[i-1]+(a[i]-a[i-1]-1)+(a[i+1]-a[i]-1);
	}
	rg int l,r;
	rg long long ans=0;
	for(rg int i=1;i<=q;i++){
		scanf("%d%d",&l,&r);
		if(l==r) ans=k-1;
		else {
			ans=sum[r-1]-sum[l];
			ans+=k-a[r-1]-1;
			ans+=a[l+1]-2;
		}
		printf("%lld\n",ans);
	}
	return 0;
}