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<cstdio>
#include<cmath>
#include<algorithm>
#define EPS 1e-6	

struct pt {
	int x, y;
	pt(){};
	pt(int _x, int _y) : x(_x), y(_y){};


	bool operator<(pt p) {
		return(x < p.x || (x == p.x && y < p.y));
	}

	bool operator==(pt p) const {
		return(x == p.x && y == p.y);
	};
};

pt points[100];


typedef std::pair<pt, pt> seg;

inline pt operator-(pt a, pt b) {
	return pt(a.x-b.x, a.y-b.y);
}

inline double cross(pt a, pt b) {
	return(a.x*b.y - a.y*b.x);
}

inline bool ccw(pt a, pt b, pt c) {
	return cross(b-a, c-a) >= 0;
}

inline double dot(pt a, pt b) {
	return (a.x*b.x + a.y*b.y);
}


void find(pt p, int numPoints) {
	double r;
	pt pa, pb;
	bool cut = false;

	// assume that there is a horizontal line ray with 
	// p as its origin in the easterly direction
	for (int i = 0; i < numPoints; ++i) {
		if (p == points[i]) {
			printf("1\n");
			return;
		}
		
		pa = points[i] - p;
		pb = points[(i+1)%numPoints] - p;

		// check that vectors are parallel && 
		// have 0 distance between the two
		// i.e. are collinear
		if (cross(pa, pb) == 0 && dot(pa, pb) <= EPS) {
			printf("1\n");
			return;
		}

		// otherwise check if p-ray cuts across ab
		if (pa.y < pb.y) {
			pt temp = pa;
			pa = pb;
			pb = temp;
		}

		// pa points more up
		// pb points more down
		// i.e. ab is a vaguely vertical interval
		if (pa.y >= 0 && pb.y < 0) {
			// p-ray cuts between a and b
			if (cross(pa,pb) > 0) cut = (!cut);
		}

				
	}
	if (cut) {
		printf("2\n");
	} else {
		printf("0\n");
	}
}

int main() {
	int numPoints, numTests, val, a, b;
	scanf("%d", &numPoints);
	for (int i = 0; i < numPoints; ++i) {
		scanf("%d %d", &a, &b);
		points[i] = pt(a, b);
	}

	scanf("%d", &numTests);
	for (int i = 0; i < numTests; ++i) {
		scanf("%d %d", &a, &b);
		find(pt(a,b), numPoints);
	}

	return 0;
}
