For a given polygon g and target points t, print "2" if g contains t, "1" if t is on a segment of g, "0" otherwise.

g is represented by a sequence of points p1, p2,..., pn where line segments connecting pi and pi+1 (1 ≤ i ≤ n-1) are sides of the polygon. The line segment connecting pn and p1 is also a side of the polygon.

Note that the polygon is not necessarily convex.

Constraints

* 3 ≤ n ≤ 100
* 1 ≤ q ≤ 1000
* -10000 ≤ xi, yi ≤ 10000
* No point of the polygon will occur more than once.
* Two sides of the polygon can intersect only at a common endpoint.

Input

The entire input looks like:


g (the sequence of the points of the polygon)
q (the number of queris = the number of target points)
1st query
2nd query
:
qth query


g is given by coordinates of the points p1,..., pn in the following format:


n
x1 y1
x2 y2
:
xn yn


The first integer n is the number of points. The coordinate of a point pi is given by two integers xi and yi. The coordinates of points are given in the order of counter-clockwise visit of them.

Each query consists of the coordinate of a target point t. The coordinate is given by two intgers x and y.

Output

For each query, print "2", "1" or "0".

Example

Input

4
0 0
3 1
2 3
0 3
3
2 1
0 2
3 2


Output

2
1
0
#include <bits/stdc++.h>
#define eps 1e-9
#define pi acos(-1)
#define nmax 200
#define f(c,a,b) for(int c=a; c<=b; c++)

using namespace std;
typedef double db;
struct P{
    db x, y;
    P(){}
    P(db x, db y) : x(x) , y(y) {}
    P operator - (P a){ return P(x-a.x, y-a.y); }
    db dot(P a){ return x*a.x+y*a.y; }
    db clen() { return sqrt(x*x+y*y); }
    db times(P a){ return x*a.y-y*a.x; }
}po[nmax];
typedef P V;
int n;

int sign(db x){ return x<-eps ? -1 : x>eps; }

int contain(P a){
    bool flag=false;
    f(i,1,n){
        P b = po[i]-a, c = po[i-1]-a;
        if( !sign(b.times(c)) && b.dot(c)<eps ) return 1;
        if(b.y < c.y) { P t=b; b=c; c=t; }
        if(c.times(b)>eps && b.y>eps && c.y<eps ) flag=!flag;
    }    
    return flag?2:0;
}

int main(){
    //freopen("owo.in","r",stdin);
    cin >> n;
    f(i,1,n) scanf("%lf%lf", &po[i].x, &po[i].y);
    po[0] = po[n];
    int q;
    cin >> q;
    while(q--) {
        P in;
        scanf("%lf%lf", &in.x, &in.y);
        printf("%d\n", contain(in));
    }
    return 0;
}
