p03685 AtCoder Regular Contest 076 - Connected?_16126
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.

Examples

Input

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


Output

YES


Input

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


Output

NO


Input

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


Output

YES


Input

1 1 2
0 0 1 1
1 0 0 1


Output

NO
#include "bits/stdc++.h"
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const LL MOD = 1000000007LL;
vector<P> v;
bool appear[100000];
int main() {
	int R, C, N;
	cin >> R >> C >> N;
	for (int i = 0; i < N; i++) {
		int x[2], y[2];
		bool ok = 1;
		for (int j = 0; j < 2; j++) {
			cin >> x[j] >> y[j];
			if (x[j] != 0 && x[j] != R&&y[j] != 0 && y[j] != C) ok = 0;
		}
		if (!ok) continue;
		for (int j = 0; j < 2; j++) {
			if (y[j] == 0) {
				v.emplace_back(x[j], i);
			}
			else if (x[j] == R) {
				v.emplace_back(R + y[j], i);
			}
			else if (y[j] == C) {
				v.emplace_back(R + C + (R - x[j]), i);
			}
			else if (x[j] == 0) {
				v.emplace_back(R + C + R + (C - y[j]), i);
			}
		}
	}
	sort(v.begin(), v.end());
	stack<int> S;
	for (int i = 0; i < v.size(); i++) {
		if (appear[v[i].second]) {
			if (S.top() != v[i].second) {
				cout << "NO" << endl;
				return 0;
			}
			S.pop();
		}
		else {
			appear[v[i].second] = 1;
			S.push(v[i].second);
		}
	}
	cout << "YES" << endl;
}