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 math import ceil, floor
from collections import defaultdict 
 
for _ in range(1):
    
    n,q,k = map(int,input().split())
    arr = list(map(int,input().split()))
    
    pp = []
    temp = []
    ee = 0
    for i in range(n):
        
        if i==0 and n>1:
            ee += arr[1]-2
            temp.append(ee)
        elif i==0 and n==1:
            ee += k-1
            temp.append(ee)
            
        elif i==n-1:
            ee += k-arr[i-1]-1
            temp.append(ee)
            
        else:
            ee += arr[i+1]-arr[i-1]-2
            temp.append(ee)
            
    
    for i in range(q):
        
        
        l,r = map(int,input().split())
    
        ans = 0
        
        if l==r:
            ans = k-1
            
        else:
            # for i in range(l-1,r):
                
            #     if i==l-1 and i!=0 and i<n-1:
            #         ans += arr[i+1]-2 
                    
            #     elif i==r-1 and i!=n-1 and i!=0:
            #         ans += k-arr[i-1]-1 
                    
            #     else:
            #         ans += temp[i]
            
            l -= 1 
            r -= 1 
            ans = (arr[l+1]-2) + (k-arr[r-1]-1) + (temp[r-1]-temp[l])
                
        print(ans)
                