Problem:
Score : 300 points

Problem Statement

Takahashi made N problems for competitive programming.
The problems are numbered 1 to N, and the difficulty of Problem i is represented as an integer d_i (the higher, the harder).

He is dividing the problems into two categories by choosing an integer K, as follows:

A problem with difficulty K or higher will be for ARCs.

A problem with difficulty lower than K will be for ABCs.

How many choices of the integer K make the number of problems for ARCs and the number of problems for ABCs the same?

Problem Statement

2 \leq N \leq 10^5

N is an even number.

1 \leq d_i \leq 10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
d_1 d_2 ... d_N

Output

Print the number of choices of the integer K that make the number of problems for ARCs and the number of problems for ABCs the same.

Sample Input 1

6
9 1 4 4 6 7

Sample Output 1

2

If we choose K=5 or 6, Problem 1, 5, and 6 will be for ARCs, Problem 2, 3, and 4 will be for ABCs, and the objective is achieved.
Thus, the answer is 2.

Sample Input 2

8
9 1 14 5 5 4 4 14

Sample Output 2

0

There may be no choice of the integer K that make the number of problems for ARCs and the number of problems for ABCs the same.

Sample Input 3

14
99592 10342 29105 78532 83018 11639 92015 77204 30914 21912 34519 80835 100000 1

Sample Output 3

42685

Solution in Python (3.4.3) (judge verdict: Runtime Error)

n,m = map(int,input().split())
a = []
mod = 10**9+7
for i in range(m):
    b = int(input())
    a.append(b)
d = [0 for i in range(n+1)]
d[0] = 0
d[1] = 1
d[2] = 2
for i in range(3,n+1):
    d[i] = d[i-1] + d[i-2]
ans = 1
if a[0] == 1:
    ans = 1
else:
    ans = ans*d[a[0]-1]%mod
for i in range(m-1):
    ans = ans*d[a[i+1]-a[i]-2]%mod
ans = ans*d[n-a[m-1]-1]%mod
print(ans)