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].
from sys import stdin
input=lambda : stdin.readline().strip()
lin=lambda :list(map(int,input().split()))
iin=lambda :int(input())
main=lambda :map(int,input().split())
from math import ceil,sqrt,factorial,log
from collections import deque
from bisect import bisect_left
mod=998244353
mod=1000000007
def gcd(a,b):
	a,b=max(a,b),min(a,b)
	while a%b!=0:
		a,b=b,a%b
	return b
def moduloinverse(a):
	return(pow(a,mod-2,mod))
def solve(we):
	n,q,k=main()
	l=lin()
	l=[1]+l
	l.append(k)
	# start=[0]*(n+1)
	mid=[0]*(n+1)
	# print(l)
	for i in range(1,n+1):
		# start[i]=l[i+1]-1
		mid[i-1]=l[i+1]-l[i-1]-2
	# print(mid)
	for i in range(1,n+1):
		mid[i]+=mid[i-1]
	# print(mid)
	for i in range(q):
		a,b=main()
		ans=0
		if a==b:
			ans=k-1
		elif b-a==1:
			ans=l[a+1]-2+k-l[b-1]-1
		else:
			ans=l[a+1]-2+k-l[b-1]-1
			# print(ans,'jkkl')
			ans+=mid[b-2]-mid[a-1]
			# print(mid[b-2]-mid[a-1],'errrrrrrr')
		print(ans)







qwe=1
# qwe=iin()
for _ in range(qwe):
	solve(_+1)