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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define FOR(i, m, n) for (ll i = m; i < (n); i++)
#define RFOR(i, m, n) for (ll i = (m - 1); i >= 0; i--)
#define REP(i, n) FOR(i, 0, n)
#define RREP(i, n) RFOR(i, n, 0)
#define ALL(v) v.begin(), v.end()
#define RALL(v) v.rbegin(), v.rend()
#define print(ele) cout << (ele) << "\n"
#define print10(ele) cout << fixed << setprecision(10) << (ele) << "\n"

using namespace std;
typedef long long ll;
typedef vector<int> ivec;
typedef vector<string> svec;
typedef vector<ll> lvec;

const int mod = 1e9 + 7;
const ll INF = 1000000000000000000LL;

int main() {
  int n, m, x;
  cin >> n >> m >> x;
  ivec a(m);
  REP(i, m) cin >> a[i];
  int ansA = 0;
  RFOR(i, x, 0) {
    REP(j, m) {
      if (i == a[j]) ansA++;
    }
    }
  int ansB = 0;
  FOR(i, x, n) {
    REP(j, m) {
      if (i == a[j]) ansB++;
    }
  }
  print(min(ansA, ansB));
  return 0;
}