Codeforces Round #512 (Div. 2）B.Vasya and Cornfield

B. Vasya and Cornfield

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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<n≤1001≤d<n≤100 ).

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.

 

题意

　　给一个矩形，判断下面的点是否在矩形内。

分析

　　判断点是否在上下两条边内，和左右两条边内就可以了。模板奉上。

 

///  author:Kissheart  ///
#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
#include<vector>
#include<stdlib.h>
#include<math.h>
#include<queue>
#include<deque>
#include<ctype.h>
#include<map>
#include<set>
#include<stack>
#include<string>
#define INF 0x3f3f3f3f
#define FAST_IO ios::sync_with_stdio(false)
const double PI = acos(-1.0);
const double eps = 1e-6;
const int MAX=1e6+10;
const int mod=1e9+7;
typedef long long ll;
using namespace std;
#define gcd(a,b) __gcd(a,b)
inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
inline ll qpow(ll a,ll b){ll r=1,t=a; while(b){if(b&1)r=(r*t)%mod;b>>=1;t=(t*t)%mod;}return r;}
inline ll inv1(ll b){return qpow(b,mod-2);}
inline ll exgcd(ll a,ll b,ll &x,ll &y){if(!b){x=1;y=0;return a;}ll r=exgcd(b,a%b,y,x);y-=(a/b)*x;return r;}
inline ll read(){ll x=0,f=1;char c=getchar();for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;for(;isdigit(c);c=getchar()) x=x*10+c-'0';return x*f;}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
double n,d;
struct Point
{
    double x;
    double y;
    Point(double x,double y)
    {
        this->x = x;
        this->y = y;
    }
};
double GetCross(Point& p1, Point& p2,Point& p)
{
    return (p2.x - p1.x) * (p.y - p1.y) -(p.x - p1.x) * (p2.y - p1.y);
}
int IsPointInMatrix(Point& p)
{
    Point p1(0,d);
    Point p2(d,0);
    Point p3(n,n-d);
    Point p4(n-d,n);
    return GetCross(p1,p2,p)*GetCross(p3,p4,p)>=0 && GetCross(p2,p3,p)*GetCross(p4,p1,p)>=0;
}
int main()
{
    int t;
    scanf("%lf%lf",&n,&d);
    scanf("%d",&t);
    while(t--)
    {
        double x,y;
        scanf("%lf%lf",&x,&y);
        Point p(x,y);
        if(IsPointInMatrix(p)) printf("YES\n");
        else printf("NO\n");
    }
    return 0;

}