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>
using namespace std;


int main() {
	const int INF = 1e8;
	int r, c, n; cin >> r >> c >> n;
	vector<pair<int, int>> v;
	for(int i = 0; i < n; ++i) {
		int x1, y1; cin >> x1 >> y1;
		int e1 = -1;
		if(x1 == 0) {
			e1 = INF - y1;
		} else if(y1 == 0) {
			e1 = INF + x1;
		} else if(x1 == r) {
			e1 = 2 * INF + y1;
		} else if(y1 == c) {
			e1 = 4 * INF - x1;
		}
		int x2, y2; cin >> x2 >> y2;
		int e2 = -1;
		if(x2 == 0) {
			e2 = INF - y2;
		} else if(y2 == 0) {
			e2 = INF + x2;
		} else if(x2 == r) {
			e2 = 2 * INF + y2;
		} else if(y2 == c) {
			e2 = 4 * INF - x2;
		}
		if(e1 != -1 and e2 != -1) {
			v.emplace_back(e1, i);
			v.emplace_back(e2, i);
		}
	}
	sort(v.begin(), v.end());
	stack<int> st;
	for(auto i : v) {
		int idx = i.second;
		if(not st.empty() and st.top() == idx) {
			st.pop();
		} else {
			st.push(idx);
		}
	}
	if(st.empty()) cout << "YES" << '\n';
	else cout << "NO" << '\n';
	return 0;
}