1485_B. Replace and Keep Sorted
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].
```python
n, q, k = map(int, input().split())
a = list(map(int, input().split()))
b, c = [], [0]
if n > 1:
	for i in range(n):
		if i == 0:
			b.append(a[i+1] - 3)
		elif i == (n-1):
			b.append(k - a[i-1] - 2)
		else: 
			b.append(a[i+1] - a[i-1] - 2)
	for i in range(n):
		c.append(c[i] + b[i])

for i in range(q):
	if n == 1:
		print(k-1)
		continue
	result = 0
	l, r = map(int, input().split())
	if l == r:
		print(k-1)
		continue
	l -= 1
	r -= 1
	result += a[l+1] + k - a[r-1] - 3
	result += c[r] - c[l+1]
	print(result)
```