Gerald plays the following game. He has a checkered field of size n × n cells, where m various cells are banned. Before the game, he has to put a few chips on some border (but not corner) board cells. Then for n - 1 minutes, Gerald every minute moves each chip into an adjacent cell. He moves each chip from its original edge to the opposite edge. Gerald loses in this game in each of the three cases:

  * At least one of the chips at least once fell to the banned cell. 
  * At least once two chips were on the same cell. 
  * At least once two chips swapped in a minute (for example, if you stand two chips on two opposite border cells of a row with even length, this situation happens in the middle of the row). 



In that case he loses and earns 0 points. When nothing like that happened, he wins and earns the number of points equal to the number of chips he managed to put on the board. Help Gerald earn the most points.

Input

The first line contains two space-separated integers n and m (2 ≤ n ≤ 1000, 0 ≤ m ≤ 105) — the size of the field and the number of banned cells. Next m lines each contain two space-separated integers. Specifically, the i-th of these lines contains numbers xi and yi (1 ≤ xi, yi ≤ n) — the coordinates of the i-th banned cell. All given cells are distinct.

Consider the field rows numbered from top to bottom from 1 to n, and the columns — from left to right from 1 to n.

Output

Print a single integer — the maximum points Gerald can earn in this game.

Examples

Input

3 1
2 2


Output

0


Input

3 0


Output

1


Input

4 3
3 1
3 2
3 3


Output

1

Note

In the first test the answer equals zero as we can't put chips into the corner cells.

In the second sample we can place one chip into either cell (1, 2), or cell (3, 2), or cell (2, 1), or cell (2, 3). We cannot place two chips.

In the third sample we can only place one chip into either cell (2, 1), or cell (2, 4).
import java.util.*;
import java.io.*;
public class a {
    static long mod = 1000000009;
    static char[][] res;
public static void main(String[] args) throws IOException
{
    //Scanner input = new Scanner(new File("input.txt"));
    //PrintWriter out = new PrintWriter(new File("output.txt"));
  input.init(System.in);
  PrintWriter out  = new PrintWriter((System.out));
  int n = input.nextInt(), m = input.nextInt();
  boolean[] rows = new boolean[n], cols = new boolean[n];
  Arrays.fill(rows, true);
  Arrays.fill(cols, true);
  for(int i = 0; i<m; i++)
  {
      cols[input.nextInt()-1] = false;
      rows[input.nextInt()-1] = false;
  }
  int count = 0;
  for(int i = 1; i<n-1; i++)
  {
      if(i > n-i-1) break;
      if(i==n-i-1)
      {
          if(rows[i] || cols[i]) count++;
          continue;
      }
      if(rows[i]) count++;
      if(rows[n-i-1]) count++;
      if(cols[i]) count++;
      if(cols[n-i-1]) count++;
  }
  out.println(count);
  out.close();
}
static long pow(long x, long p)
{
    if(p==0) return 1;
    if((p&1) > 0)
    {
        return (x*pow(x, p-1))%mod;
    }
    long sqrt = pow(x, p/2);
    return (sqrt*sqrt)%mod;
}
static long gcd(long a, long b)
{
    if(b==0) return a;
    return gcd(b, a%b);
}
static class input {
  static BufferedReader reader;
  static StringTokenizer tokenizer;

  /** call this method to initialize reader for InputStream */
  static void init(InputStream input) {
      reader = new BufferedReader(
                   new InputStreamReader(input) );
      tokenizer = new StringTokenizer("");
  }

  /** get next word */
  static String next() throws IOException {
      while ( ! tokenizer.hasMoreTokens() ) {
          //TODO add check for eof if necessary
          tokenizer = new StringTokenizer(
                 reader.readLine() );
      }
      return tokenizer.nextToken();
  }

  static int nextInt() throws IOException {
      return Integer.parseInt( next() );
  }
  
  static double nextDouble() throws IOException {
      return Double.parseDouble( next() );
  }
  static long nextLong() throws IOException {
      return Long.parseLong( next() );
  }
  static String nextLine() throws IOException {
    return reader.readLine();
  }
}
/*
Sum Interval Tree - uses O(n) space
Updates and queries over a range of values in log(n) time
Note: If querying all elements, use difference array instead.
*/
static class IT
{
    int[] left,right, val, a, b;
    IT(int n)
    {
        left = new int[4*n];
        right = new int[4*n];
        val = new int[4*n];
        a = new int[4*n];
        b = new int[4*n];
        init(0,0, n);
    }
    int init(int at, int l, int r)
    {
        a[at] = l;
        b[at] = r;
        if(l==r)
            left[at] = right [at] = -1;
        else 
        {
            int mid = (l+r)/2;
            left[at] = init(2*at+1,l,mid);
            right[at] = init(2*at+2,mid+1,r);
        }
        return at++;
    }
    //return the sum over [x,y]
    int get(int x, int y)
    {
        return go(x,y, 0);
    }
    int go(int x,int y, int at)
    {
        if(at==-1) return 0;
        if(x <= a[at] && y>= b[at]) return val[at];
        if(y<a[at] || x>b[at]) return 0;
        return go(x, y, left[at]) + go(x, y, right[at]);
    }
    //add v to elements x through y
    void add(int x, int y, int v)
    {
        go3(x, y, v, 0);
    }
    void go3(int x, int y, int v, int at)
    {
        if(at==-1) return;
        if(y < a[at] || x > b[at]) return;
        val[at] += (Math.min(b[at], y) - Math.max(a[at], x) + 1)*v;
        go3(x, y, v, left[at]);
        go3(x, y, v, right[at]);
    }
}
}