cf 512 div2 B. Vasya and Cornfield

Vasya owns a cornfield which can be defined with two integers nn and dd. The cornfield can be represented as rectangle with vertices having Cartesian coordinates (0,d),(d,0),(n,n−d)(0,d),(d,0),(n,n−d) and (n−d,n)(n−d,n).

An example of a cornfield with n=7n=7 and d=2d=2.

Vasya also knows that there are mm grasshoppers near the field (maybe even inside it). The ii-th grasshopper is at the point (xi,yi)(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 nn and dd (1≤d

The second line contains a single integer mm (1≤m≤1001≤m≤100) — the number of grasshoppers.

The ii-th of the next mm lines contains two integers xixi and yiyi (0≤xi,yi≤n0≤xi,yi≤n) — position of the ii-th grasshopper.

Output

Print mm lines. The ii-th line should contain "YES" if the position of the ii-th grasshopper lies inside or on the border of the cornfield. Otherwise the ii-th line should contain "NO".

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

Examples

input

Copy

7 2
4
2 4
4 1
6 3
4 5

output

Copy

YES
NO
NO
YES

input

Copy

8 7
4
4 4
2 8
8 1
6 1

output

Copy

YES
NO
YES
YES

Note

The cornfield from the first example is pictured above. Grasshoppers with indices 11 (coordinates (2,4)(2,4)) and 44 (coordinates (4,5)(4,5)) are inside the cornfield.

The cornfield from the second example is pictured below. Grasshoppers with indices 11 (coordinates (4,4)(4,4)), 33 (coordinates (8,1)(8,1)) and 44(coordinates (6,1)(6,1)) are inside the cornfield.

给出4个点组成一个矩形，给出n个查询点，问这些点是否在矩形内。

第一次看挺难，给出的4个点，可以推一下。发现斜率都是1或-1.

然后就判断就好了

#pragma GCC optimize(2)
#include
#include
#include
#include
using namespace std;
const int maxn = 500;
const int inf = 0x3f3f3f3f;
typedef long long ll;
int a[maxn];
int main()
{
	//freopen("C://input.txt", "r", stdin);
	int n;
	int flag = 0;
	scanf("%d", &n);
	for (int i = 0; i < n; i++)
	{
		int a;
		scanf("%d", &a);
		if (a == 1)
		{
			flag = 1;
		}
	}
	if (flag == 1)
	{
		printf("HARD\n");
	}
	else
	{
		printf("EASY\n");
	}
	return 0;
}