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].
# cook your dish here
import os,sys;from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
    newlines = 0
    def __init__(self, file):
        self._fd = file.fileno();self.buffer = BytesIO();self.writable = "x" in file.mode or "r" not in file.mode;self.write = self.buffer.write if self.writable else None
    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:break
            ptr = self.buffer.tell();self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()
    def readline(self):
        while self.newlines == 0:b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE));self.newlines = b.count(b"\n") + (not b);ptr = self.buffer.tell();self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()
    def flush(self):
        if self.writable:os.write(self._fd, self.buffer.getvalue());self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file);self.flush = self.buffer.flush;self.writable = self.buffer.writable;self.write = lambda s: self.buffer.write(s.encode("ascii"));self.read = lambda: self.buffer.read().decode("ascii");self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
try:sys.stdin,sys.stdout=open('in.txt','r'),open('out.txt','w')
except:pass
ii1=lambda:int(sys.stdin.readline().strip()) # for interger
is1=lambda:sys.stdin.readline().strip() # for str
iia=lambda:list(map(int,sys.stdin.readline().strip().split())) # for List[int]
isa=lambda:sys.stdin.readline().strip().split() # for List[str]
mod=int(1e9 + 7);from collections import *;from math import *
# abc = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
# sys.setrecursionlimit(500000)
###################### Start Here ######################










def main():
    n,q,k = iia()
    arr = iia()
    res = []
    if n>1:
        for i in range(n):
            if i==0:
                res.append(arr[1]-2)
            elif i==n-1:
                res.append(k-arr[i-1]-1)
            else:
                res.append(abs(arr[i+1]-arr[i-1]-2))
    # print(res)
    if n>1:
        new = [res[0]]
        for i in range(1,len(res)):
            new.append(res[i]+new[-1])

    for i in range(q):
        l,r = iia();l-=1;r-=1;
        if n==1:
            print(k-1)
        else:
            if l==r:
                print(k-1)
            elif l==0:
                ans = 0
                ans += max(arr[l+1]-2,0)
                ans += max(k-arr[r-1]-1,0)
                if r-l==2:
                    ans+=res[l+1]
                elif r-l>2:
                    ans+=new[r-1]-new[l]
                print(ans)
            else:
                ans = 0
                ans+= max(arr[l+1]-2,0)
                ans += max(k-arr[r-1]-1,0)
                if r-l==2:
                    ans+=res[l+1]
                elif r-l>2:
                    ans+= new[r-1]-new[l]
                print(ans)






if __name__ == '__main__':
    main()
