A. Vika and Her Friends

Vika and her friends went shopping in a mall, which can be represented as a rectangular grid of rooms with sides of length $n$ and $m$. Each room has coordinates $(a, b)$, where $1 \le a \le n, 1 \le b \le m$. Thus we call a hall with coordinates $(c, d)$ a neighbouring for it if $|a - c| + |b - d| = 1$.

Tired of empty fashion talks, Vika decided to sneak away unnoticed. But since she hasn't had a chance to visit one of the shops yet, she doesn't want to leave the mall. After a while, her friends noticed Vika's disappearance and started looking for her.

Currently, Vika is in a room with coordinates $(x, y)$, and her $k$ friends are in rooms with coordinates $(x_1, y_1)$, $(x_2, y_2)$, ... $, (x_k, y_k)$, respectively. The coordinates can coincide. Note that all the girls must move to the neighbouring rooms.

Every minute, first Vika moves to one of the adjacent to the side rooms of her choice, and then each friend (seeing Vika's choice) also chooses one of the adjacent rooms to move to.

If at the end of the minute (that is, after all the girls have moved on to the neighbouring rooms) at least one friend is in the same room as Vika, she is caught and all the other friends are called.

Tell us, can Vika run away from her annoying friends forever, or will she have to continue listening to empty fashion talks after some time?

Input

Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 100$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains three integers $n$, $m$, $k$ ($1 \le n, m, k \le 100$) — the sizes of the mall and the number of Vika's friends.

The second line of each test case contains a pair of integers $x$ and $y$ ($1 \le x \le n$, $1 \le y \le m$) — the coordinates of the room where Vika is.

Each of the next $k$ lines of each test case contains a pair of integers $x_i$ and $y_i$ ($1 \le x_i \le n$, $1 \le y_i \le m$) — the coordinates of the room where the $i$-th friend is.

Output

For each test case, output "YES" if Vika can run away from her friends forever, otherwise output "NO".

You can output each letter in any case (lowercase or uppercase). For example, the strings "yEs", "yes", "Yes", and "YES" will be accepted as a positive answer.

Example

input

output

YES
NO
YES
NO
YES
YES

Note

In the first test case, the friend will never catch up with Vika, because Vika can always move to the room diagonally opposite to the one where the friend is.

In the second test case, no matter where Vika goes, each of her friends can catch her after the first move.

In the third test case, Vika and her friend will always be in different halls.

解题思路

　　不妨先把$n \times m$的棋盘染成黑白相间的形式，如果$x+y$为偶数则染白色，否则染黑色，如下图：

　　先给出结论，如果所有的$(x_i, y_i)$都与$(x, y)$的格子颜色不同，那么在移动的过程中任意一个$(x_i, y_i)$都无法与$(x, y)$相遇，否则（即存在至少一个$(x_i, y_i)$的格子颜色与$(x, y)$相同）必然存在一个$(x_i, y_i)$与$(x, y)$相遇。

　　首先如果$(x_i, y_i)$与$(x, y)$的格子颜色不同，那么在一次移动后格子颜色还是不同的（白变黑，黑变白），因此这两个格子永远无法相遇。

　　然后再证明如果$(x_i, y_i)$与$(x, y)$的格子颜色相同，那么在移动的过程中$(x_i, y_i)$一定会与$(x, y)$相遇。定义$(x_i, y_i)$与$(x, y)$相对位置的面积为这两个点所构成的矩形面积，如下图：

　　当$(x, y)$移动后，$(x_i, y_i)$总是沿着使得矩形面积减少或不变的方向移动，由于棋盘的大小有限，因此不会存在总是使得面积不变的移动方案，即必然会存在使得面积减少的方案，因此在移动的过程中矩形面积呈非递增，最后减少到$0$，即两个格子会相遇。

　　AC代码如下：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 void solve() {
 5     int n, x, y;
 6     scanf("%*d %*d %d %d %d", &n, &x, &y);
 7     bool flag = true;
 8     while (n--) {
 9         int a, b;
10         scanf("%d %d", &a, &b);
11         if ((x + y & 1) == (a + b & 1)) flag = false;
12     }
13     printf("%s\n", flag ? "YES" : "NO");
14 }
15 
16 int main() {
17     int t;
18     scanf("%d", &t);
19     while (t--) {
20         solve();
21     }
22     
23     return 0;
24 }