Problem:
Score : 700 points

Problem Statement

Snuke is playing a puzzle game.
In this game, you are given a rectangular board of dimensions R × C, filled with numbers. Each integer i from 1 through N is written twice, at the coordinates (x_{i,1},y_{i,1}) and (x_{i,2},y_{i,2}).

The objective is to draw a curve connecting the pair of points where the same integer is written, for every integer from 1 through N.
Here, the curves may not go outside the board or cross each other.

Determine whether this is possible.

Constraints

1 ≤ R,C ≤ 10^8

1 ≤ N ≤ 10^5

0 ≤ x_{i,1},x_{i,2} ≤ R(1 ≤ i ≤ N)

0 ≤ y_{i,1},y_{i,2} ≤ C(1 ≤ i ≤ N)

All given points are distinct.

All input values are integers.

Input

Input is given from Standard Input in the following format:

R C N
x_{1,1} y_{1,1} x_{1,2} y_{1,2}
:
x_{N,1} y_{N,1} x_{N,2} y_{N,2}

Output

Print YES if the objective is achievable; print NO otherwise.

Sample Input 1

4 2 3
0 1 3 1
1 1 4 1
2 0 2 2

Sample Output 1

YES

The above figure shows a possible solution.

Sample Input 2

2 2 4
0 0 2 2
2 0 0 1
0 2 1 2
1 1 2 1

Sample Output 2

NO

Sample Input 3

5 5 7
0 0 2 4
2 3 4 5
3 5 5 2
5 5 5 4
0 3 5 1
2 2 4 4
0 5 4 1

Sample Output 3

YES

Sample Input 4

1 1 2
0 0 1 1
1 0 0 1

Sample Output 4

NO

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

#include<bits/stdc++.h>
#define maxn 200005
using namespace std;
typedef long long LL;
int read()
{
    char c=getchar();int f=1,sum=0;
    while(c<'0' || c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0' && c<='9'){sum=sum*10+c-'0';c=getchar();}
    return sum*f;
}
int r,c,n;
struct point{
	int id,x,y;
}p[maxn];
int tot,top;
int s[maxn];
bool onedge(point x){return !x.x||!x.y||x.x==r||x.y==c;}
int calc(point x)
{
	if(x.y==0) return x.x;
    if(x.x==r) return r+x.y;
    if(x.y==c) return 2*r+c-x.x;
	if(x.x==0) return 2*r+2*c-x.y;
}
bool cmp(point x,point y){return calc(x)<calc(y);}
int main()
{
	r=read();c=read();n=read();
	for(int i=1;i<=n;i++)
	{
		point x1,x2;
		x1.id=x2.id=i;
		x1.x=read();x1.y=read();
		x2.x=read();x2.y=read();
		if(onedge(x1) && onedge(x2)) p[++tot]=x1,p[++tot]=x2;
	}
	sort(p+1,p+1+tot,cmp);
	for(int i=1;i<=tot;i++)
	{
		if(top && s[top]==p[i].id) top--;
		else s[++top]=p[i].id;
	}
	puts(top?"NO":"YES");
	return 0;
}