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 = stdin.readline
_int, _range = int, range
def solution():
    n, q, k = [_int(i) for i in _input().split()]
    arr = [_int(i) for i in _input().split()]
    cum = [0]*n
    for i in _range(n-1):
        cum[i] = arr[i+1] - arr[i] - 1
    ra = [0]*(n+1)
    ra[0] = 0
    for i in _range(1, n+1):
        ra[i] = cum[i-1] + ra[i-1]
    for i in _range(q):
        l, r = [_int(i) for i in _input().split()]
        l -= 1
        r -= 1
        if l == r:
            print(k-1)
        else:
            ans = arr[l+1] - 2 + (k - arr[r-1]) - 1
            if r - l > 1:
                otr = (ra[r] - ra[l])
                ans += otr*2
                ans -= cum[l] + cum[r-1]
            print(ans)

solution()