Problem:
Score : 200 points

Problem Statement

There are N + 1 squares arranged in a row, numbered 0, 1, ..., N from left to right.

Initially, you are in Square X.
You can freely travel between adjacent squares. Your goal is to reach Square 0 or Square N.
However, for each i = 1, 2, ..., M, there is a toll gate in Square A_i, and traveling to Square A_i incurs a cost of 1.
It is guaranteed that there is no toll gate in Square 0, Square X and Square N.

Find the minimum cost incurred before reaching the goal.

Constraints

1 \leq N \leq 100

1 \leq M \leq 100

1 \leq X \leq N - 1

1 \leq A_1 < A_2 < ... < A_M \leq N

A_i \neq X

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M X
A_1 A_2 ... A_M

Output

Print the minimum cost incurred before reaching the goal.

Sample Input 1

5 3 3
1 2 4

Sample Output 1

1

The optimal solution is as follows:

First, travel from Square 3 to Square 4. Here, there is a toll gate in Square 4, so the cost of 1 is incurred.

Then, travel from Square 4 to Square 5. This time, no cost is incurred.

Now, we are in Square 5 and we have reached the goal.

In this case, the total cost incurred is 1.

Sample Input 2

7 3 2
4 5 6

Sample Output 2

0

We may be able to reach the goal at no cost.

Sample Input 3

10 7 5
1 2 3 4 6 8 9

Sample Output 3

3

Solution in Python (3.4.3) (judge verdict: Accepted)

# Author: cr4zjh0bp
# Created: Thu Mar 12 07:52:52 UTC 2020
import sys
 
stdin = sys.stdin
inf = 1 << 60
mod = 1000000007
 
ni = lambda: int(ns())
nin = lambda y: [ni() for _ in range(y)]
na = lambda: list(map(int, stdin.readline().split()))
nan = lambda y: [na() for _ in range(y)]
nf = lambda: float(ns())
nfn = lambda y: [nf() for _ in range(y)]
nfa = lambda: list(map(float, stdin.readline().split()))
nfan = lambda y: [nfa() for _ in range(y)]
ns = lambda: stdin.readline().rstrip()
nsn = lambda y: [ns() for _ in range(y)]
ncl = lambda y: [list(ns()) for _ in range(y)]
nas = lambda: stdin.readline().split()

N, M, X = na()
X -= 1
a = na()
b = [0] * N
for i in range(M):
    b[a[i] - 1] = 1

l, r = 0, 0
for i in range(N):
    if i < X:
        l += b[i]
    else:
        r += b[i]

print(min(l, r))