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 <iostream>
#include <algorithm>
#include <map>
#include <vector>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;
ll h,w,n;
ll len(ll x,ll y){
  if(x==0)
    return y;
  if(y==h)
    return h+x;
  if(x==w)
    return h+w+(h-y);
  if(y==0)
    return h+w+h+(w-x);
  return -1;
}

vector<P> pos;
map<ll,int> zahyou;
vector<ll> X;
int tbl[300001];
int main(){
  cin >> h >> w >> n;
  for(int i=0;i<n;++i){
    ll x1,y1,x2,y2;
    cin >> y1 >> x1 >> y2 >> x2;
    ll left = len(x1,y1);
    ll right = len(x2,y2);
    if(right < left)
      swap(right,left);
    if(left>=0)
      pos.push_back(P(left,right));
  }
  sort(pos.begin(), pos.end());

  for(P p:pos){
    X.push_back(p.first);
    X.push_back(p.second);
  }
  sort(X.begin(), X.end());
  for(int i=0;i<X.size();++i){
    zahyou[X[i]]=i;
  }
  for(P p:pos){
    tbl[zahyou[p.first]]++;
    tbl[zahyou[p.second]+1]--;
  }
  for(int i=1;i<=300000;++i){
    tbl[i]+=tbl[i-1];
  }

  bool ok = true;
  for(P p:pos){
    if(tbl[zahyou[p.first]]!=tbl[zahyou[p.second]])
      ok=false;
  }

  if(ok)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
 
  return 0;
}