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>
using namespace std;

typedef long double ld;

const ld EPS = 1e-9;
const ld INF = 1e10;
const ld PI = M_PI;

struct Point{
  ld x, y;
  Point(ld x, ld y):x(x), y(y){}
  Point(){}
};

Point operator+(const Point &a, const Point &b){
  return Point(a.x + b.x, a.y + b.y);
}
Point operator-(const Point &a, const Point &b){
  return Point(a.x - b.x, a.y - b.y);
}
Point operator*(const Point &a, const ld b){
  return Point(a.x * b, a.y * b);
}
Point operator*(const Point &a, const Point &b){
  return Point(a.x*b.x-a.y*b.y, a.x*b.y+a.y*b.x);
}
Point operator/(const Point &a, const ld b){
  return Point(a.x / b, a.y / b);
}

ld cross(const Point &a, const Point &b){
  return a.x*b.y - a.y*b.x;
}

ld dot(const Point &a, const Point &b){
  return a.x*b.x + a.y*b.y;
}

ld norm(const Point &a){
  return dot(a, a);
}

struct Line:vector<Point>{
  Line(Point a = Point(0, 0), Point b = Point(0, 0)){
    this->push_back(a);
    this->push_back(b);
  }
};

typedef vector<Point> Polygon;

int ccw(Point a, Point b, Point c){
  b = b - a;
  c = c - a;
  if(cross(b, c) > EPS) return +1; // 反時計周り
  if(cross(b, c) < -EPS) return -1; // 時計周り
  if(dot(b, c) < 0) return +2; // c -- a -- b がこの順番に一直線上
  if(norm(b) < norm(c)) return -2; // a -- b -- c がこの順番に一直線上
  return 0; // a -- c -- b が一直線上
}

Point next(const Polygon &a, int x){
  return a[(x+1)%a.size()];
}

Point prev(const Polygon &a, int x){
  return a[(x-1+a.size()) % a.size()];
}

bool is_convex(const Polygon &poly){
  for(int i = 0 ; i < (int)poly.size() ; i++){
    if(ccw(prev(poly, i), poly[i], next(poly, i)) == -1) return false;
  }
  return true;
}

int contain(const Polygon &poly, const Point &p){
  bool in = false;
  for(int i = 0 ; i < (int)poly.size() ; i++){
    Point a = poly[i] - p;
    Point b = next(poly, i) - p;
    if(a.y > b.y) swap(a, b);
    if(a.y <= 0 && 0 < b.y && cross(a, b) < 0) in = !in;
    if(cross(a, b) == 0 && dot(a, b) <= 0) return 1;
  }
  return in ? 2 : 0;
}

int main(){  
  int q;
  cin >> q;
  Polygon poly;
  while(q--){
    Point p;
    cin >> p.x >> p.y;
    poly.push_back(p);
  }
  cin >> q;
  while(q--){
    Point p;
    cin >> p.x >> p.y;
    cout << contain(poly, p) << endl;    
  }
  return 0;
}