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].
#####################################   
import atexit, io, sys, collections, math, heapq,fractions
buffer = io.BytesIO() 
sys.stdout = buffer
@atexit.register 
def write(): sys.__stdout__.write(buffer.getvalue())   
####################################


def f(u, v, ais, k, precompute):
	t = 0
	if u == v:
		return (k-1) if ais[u] <= k else 0

	if ais[v] > k:
		return 0

	t += precompute[v-1] if v-1>= 0 else 0
	t -= precompute[u] if u >= 0 else 0
	for cur in [u,v]:
		if ais[cur] > k: return 0
		diff = ais[cur + 1] if (cur != v) else (k+1)
		diff -= ais[cur - 1] if (cur != u) else 0
		t += diff +1 - 2 - 1
		
	return t
n,q,k = map(int, raw_input().split())
ais = map(int, raw_input().split())
precompute = [0 for ii in range(len(ais))]
for cur in range(0, len(ais)):
	diff = ais[cur +1] if (cur+1 < len(ais)) else (k+1)
	diff -= ais[cur -1] if (cur -1 >= 0) else 0
	t = diff +1 - 2 -1
	precompute[cur] = t
	if cur: precompute[cur] += precompute[cur-1]

for o in range(q):
	u,v = map(int, raw_input().split())
	u-=1
	v-=1
	print f(u,v,ais,k, precompute)
