【题解】codeforces1058B[Codeforces Round #512 Div.2 B]Vasya and Cornfield 数学知识

Description

Vasya owns a cornfield which can be defined with two integers n and d. The cornfield can be represented as rectangle with vertices having Cartesian coordinates (0,d),(d,0),(n,n−d) and (n−d,n).
【题解】codeforces1058B[Codeforces Round #512 Div.2 B]Vasya and Cornfield 数学知识
An example of a cornfield with n=7 and d=2.
Vasya also knows that there are m grasshoppers near the field (maybe even inside it). The i-th grasshopper is at the point (xi,yi). Vasya does not like when grasshoppers eat his corn, so for each grasshopper he wants to know whether its position is inside the cornfield (including the border) or outside.

Help Vasya! For each grasshopper determine if it is inside the field (including the border).

Input

The first line contains two integers n and d (1≤d<n≤100).

The second line contains a single integer m (1≤m≤100) — the number of grasshoppers.

The i-th of the next m lines contains two integers xi and yi (0≤xi,yi≤n) — position of the i-th grasshopper.

Output

Print m lines. The i-th line should contain “YES” if the position of the i-th grasshopper lies inside or on the border of the cornfield. Otherwise the i-th line should contain “NO”.

You can print each letter in any case (upper or lower).

Examples

Input

7 2
4
2 4
4 1
6 3
4 5

Output

YES
NO
NO
YES

Input

8 7
4
4 4
2 8
8 1
6 1

Output

YES
NO
YES
YES

Note

The cornfield from the first example is pictured above. Grasshoppers with indices 1 (coordinates (2,4)) and 4 (coordinates (4,5)) are inside the cornfield.

The cornfield from the second example is pictured below. Grasshoppers with indices 1 (coordinates (4,4)), 3 (coordinates (8,1)) and 4 (coordinates (6,1)) are inside the cornfield.
【题解】codeforces1058B[Codeforces Round #512 Div.2 B]Vasya and Cornfield 数学知识


一道线性规划题。p(x,y)\forall p(x,y),点 pp 在四条直线围成的矩形中,当且仅当 x,yx,y 满足以下约束条件:
{x+yd0x+y2n+d0xyd0xy+d0\begin{cases}x+y-d\geq0\\x+y-2n+d\leq0\\x-y-d\leq0\\x-y+d\geq0\end{cases}
然后读入每个点判断就行了。

#include<cstdio>
int main()
{
	//freopen("in.txt","r",stdin);
	int n,d,x,y,m;
	scanf("%d%d%d",&n,&d,&m);
	while(m--)
	{
		scanf("%d%d",&x,&y);
		if(x+y-d<0||x+y-2*n+d>0||x-y-d>0||x-y+d<0)puts("NO");
		else puts("YES");
	}
	return 0;
}

总结

我不碰数学好多年