p03685 AtCoder Regular Contest 076 - Connected?
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
```python
def ri(): return int(input())
def rli(): return list(map(int, input().split()))
def rls(): return list(input())
def pli(a): return "".join(list(map(str, a)))


R, C, N = rli()
lis = []
idx = 0
i = 0
for j in range(N):
    x1, y1, x2, y2 = rli()
    if((x1 % R == 0 or y1 % C == 0) and (x2 % R == 0 or y2 % C == 0)):
        i += 1
        for _ in range(2):
            if(y1 == 0):
                lis.append([x1, i])
            elif(x1 == R):
                lis.append([R+y1, i])
            elif(y1 == C):
                lis.append([2*R+C-x1, i])
            elif(x1 == 0):
                lis.append([2*R+2*C-y1, i])
            x1 = x2
            y1 = y2
lis.sort()
stack = []
exist = [False for _ in range(2*i)]
ans = True
for k in range(2*i):
    if(exist[lis[k][1]]):
        if(lis[k][1] != stack.pop()):
            ans = False
            break
        exist[lis[k][1]] = False
    else:
        stack.append(lis[k][1])
        exist[lis[k][1]] = True
if(ans):
    print("YES")
else:
    print("NO")

```