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 deque
import sys 


#==============To chaliye shuru krte he ====================#
n,q,k=map(int,input().split())
li=deque(map(int,input().split()))
if len(li)>1:
    pre=[0]*n
    pre[0]=li[1]-2
    for i in range(1,n-1):
        pre[i]=li[i+1]-li[i-1]-2
    pre[-1]=li[-2]-1
    for i in range(1,n):
        pre[i]+=pre[i-1]

for i in range(q):
    l,r=map(int,input().split())
    ans=0
    if l==r:
        ans+=li[l-1]-1+k-li[l-1]
    else:
        ans=pre[r-2]-pre[l-1]
        ans+=li[l]-2
        ans+=k-li[r-2]-1
        
    print(ans)
# for i in range(q):
#     a,b=map(int,input().split())
#     li1=deque()
#     ans=0
#     for i in range(a-1,b):
#          li1.append(li[i])
#     tr=False
#     if a==b:
#         ans+=li[a-1]-1
#         ans+=k-li[a-1]
#         tr=True
#     else:
#         ans+=li[a]-2
#     #P(ans)
#         for i in range(a,b-1):
#             ans+=li[i+1]-li[i-1]-2
#             #P(ans)
#         #print(li[b-2])
#         ans+=k-li[b-2]-1
        
#     print(ans)