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 __future__ import division
from sys import stdin, stdout
from collections import *

rstr = lambda: stdin.readline().strip()
rstrs = lambda: [str(x) for x in stdin.readline().split()]
rstr_2d = lambda n: [rstr() for _ in range(n)]
rint = lambda: int(stdin.readline())
rints = lambda: [int(x) for x in stdin.readline().split()]
rint_2d = lambda n: [rint() for _ in range(n)]
rints_2d = lambda n: [rints() for _ in range(n)]
pr = lambda args, sep: stdout.write(sep.join(map(str, args)) + '\n')
ceil1, out = lambda a, b: (a + b - 1) // b, []

n, q, k = rints()
a = rints()
qur, mem = rints_2d(q), [0] * n

for i in range(1, n - 1):
    mem[i] = a[i + 1] - a[i - 1] - 2

mem1 = [0, 0]
for i in range(1, len(mem)):
    mem1.append(mem1[-1] + mem[i])

# print(mem1, mem)
for l, r in qur:
    out.append(0)
    l1, r1 = 0, 0
    if l == r:
        out[-1] += k - 1
    else:
        l1 += (a[l] - a[l - 1] - 1) + (a[l - 1] - 1)
        r1 += k - a[r - 2] - 1
        out[-1] += l1 + r1 + (mem1[r - 1] - mem1[l])

pr(out, '\n')
