White Cloud has a square of n×n from (1,1) to (n,n).

White Rabbit wants to put in several cars. Each car will start moving at the same time and move from one side of one row or one line to the other. All cars have the same speed. If two cars arrive at the same time and the same position in a grid or meet in a straight line, both cars will be damaged.

White Cloud will destroy the square m times. In each step White Cloud will destroy one grid of the square(It will break all m grids before cars start).Any car will break when it enters a damaged grid.

White Rabbit wants to know the maximum number of cars that can be put into to ensure that there is a way that allows all cars to perform their entire journey without damage.

(update: all cars should start at the edge of the square and go towards another side, cars which start at the corner can choose either of the two directions)

For example, in a 5×5 square

legal

illegal (These two cars will collide at (4,4))

illegal (One car will go into a damaged grid)

Input
The first line of input contains two integers n and m(n≤100000,m≤100000)
For the next m lines,each line contains two integers x,y(1≤x,y≤n), denoting the grid which is damaged by White Cloud.

Output
Print a number,denoting the maximum number of cars White Rabbit can put into.

Example
Input
2 0
Output
4
Note

由note的图片提示可知，当没危险地区时的最大车辆为2*n
每次减去危险地区两个方向不能放的车辆数就行
最后注意一下n为奇数时，矩阵中间十字线，只能放一辆车，需要再减去1

#include <bits/stdc++.h>
using namespace std;
int l[100100],r[100100],x,y,ans=0;
int main() 
{
    int n,m,x,y;
    scanf("%d%d",&n,&m);
    ans=2*n;
    while(m--)
    {
    	cin>>x>>y;
    	if(!l[x]) l[x]=1,ans--;
    	if(!r[y]) r[y]=1,ans--; 
	}
	if((n&1)&&!(l[n/2+1]||r[n/2+1])) ans--;
    printf("%d",ans);
    return 0;
}