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 <cmath>
#include <vector>
#include <algorithm>
using namespace std;

#define equals(a, b) (fabs((a)-(b)) < EPS)
const double EPS = (1e-10);

class Point {
public:
    double x, y;

    Point(double x = 0, double y = 0): x(x), y(y) {}

    Point operator + (Point p) {
        return Point(x+p.x, y+p.y);
    }
    Point operator - (Point p) {
        return Point(x-p.x, y-p.y);
    }
    Point operator * (double a) {
        return Point(x*a, y*a);
    }
    Point operator / (double a) {
        return Point(x/a, y/a);
    }

    double abs() {
        return sqrt(norm());
    }
    double norm() {
        return x*x + y*y;
    }

    bool operator < (const Point &p) const {
        return x != p.x ? x < p.x : y < p.y;
    }

    bool operator == (const Point &p) const {
        return fabs(x-p.x) < EPS && fabs(y-p.y) < EPS;
    }
};

typedef Point Vector;
// 内積
double dot(Vector a, Vector b) {
    return a.x*b.x + a.y*b.y;
}
// 外積
double cross(Vector a, Vector b) {
    return a.x*b.y - a.y*b.x;
}

typedef vector<Point> Polygon;  // 多角形

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, 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(void) {
    int n; cin >> n;
    int x, y;

    Polygon g;
    for (int i = 0; i < n; i++) {
        cin >> x >> y;
        g.push_back(Point(x, y));
    }

    cin >> n;
    for (int i = 0; i < n; i++) {
        cin >> x >> y;
        cout << contains(g, Point(x, y)) << endl;
    }

    return 0;
}