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 collections import defaultdict,deque
from sys import stdin,stdout
from bisect import bisect_left
from math import gcd
print=stdout.write
'''
t=int(stdin.readline().strip())
for _ in range(t):'''
    #n=int(stdin.readline().strip())
n,m,k=map(int,stdin.readline().split())
ll=list(map(int,stdin.readline().split()))
l=[0]
l.extend(ll)
l.append(k+1)
ans=[0]*(n+2)
for i in range(1,n+1):
    temp=0
    z=l[i]-l[i-1]-1
    if z>0:
        temp+=z
    z=l[i+1]-l[i]-1
    if z>0:
        temp+=z
    ans[i]=temp
#print(ans)
for i in range(1,n+2):
    ans[i]+=ans[i-1]
#print(ans)
for i in range(m):
    x,y=map(int,stdin.readline().split())
    temp=ans[y-1]-ans[x]
    temp+=(l[x]-1)+(k-l[y])+(l[x+1]-l[x]-1)+(l[y]-l[y-1]-1)
    print(str(temp)+'\n')
    
    
    
    
    
    











