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 <iostream>
#include <complex>
#include <utility>
#include <vector>
using namespace std;

typedef complex<double> Point, Vector;
typedef pair<Point, Point> Segment, Line;
typedef vector<Point> Polygon;

#define X real()
#define Y imag()
#define EPS (1e-10)
#define equals(a, b) (fabs((a) - (b)) < EPS)

double dot(Vector a, Vector b){
    return a.X * b.X + a.Y * b.Y;
}

double cross(Vector a, Vector b){
    return a.X * b.Y - a.Y * b.X;
}

double norm(Vector a){
    return a.X * a.X + a.Y * a.Y;
}

double abs(Vector a){
    return sqrt(norm(a));
}

int contains(Polygon g, Point p){
    int n = g.size();
    bool x = false;
    for(int i=0; i<n; i++){
        Point a = g[i] - p;
        Point b = g[(i + 1) % n] - p;
        if(abs(cross(a, b)) < EPS && dot(a, b) < EPS) return 1;
        if(a.Y > b.Y) swap(a, b);
        if(a.Y < EPS && EPS < b.Y && cross(a, b) > EPS) x = !x;
    }

    return (x ? 2 : 0);
}

int main(){
    int g; cin>>g;
    Polygon poly;
    for(int i=0; i<g; i++){
        double a, b; cin>>a>>b;
        poly.emplace_back(a, b);
    }

    int n; cin>>n;
    for(int i=0; i<n; i++){
        double a, b; cin>>a>>b;
        Point p1 = {a, b};

        cout << contains(poly, p1) << endl;
    }
}
