Problem: Maria participates in a bicycle race.

The speedway takes place on the shores of Lake Lucerne, just repeating its contour. As you know, the lake shore consists only of straight sections, directed to the north, south, east or west.

Let's introduce a system of coordinates, directing the Ox axis from west to east, and the Oy axis from south to north. As a starting position of the race the southernmost point of the track is selected (and if there are several such points, the most western among them). The participants start the race, moving to the north. At all straight sections of the track, the participants travel in one of the four directions (north, south, east or west) and change the direction of movement only in bends between the straight sections. The participants, of course, never turn back, that is, they do not change the direction of movement from north to south or from east to west (or vice versa).

Maria is still young, so she does not feel confident at some turns. Namely, Maria feels insecure if at a failed or untimely turn, she gets into the water. In other words, Maria considers the turn dangerous if she immediately gets into the water if it is ignored.

Help Maria get ready for the competition — determine the number of dangerous turns on the track.

Input

The first line of the input contains an integer n (4 ≤ n ≤ 1000) — the number of straight sections of the track.

The following (n + 1)-th line contains pairs of integers (xi, yi) ( - 10 000 ≤ xi, yi ≤ 10 000). The first of these points is the starting position. The i-th straight section of the track begins at the point (xi, yi) and ends at the point (xi + 1, yi + 1).

It is guaranteed that:

  * the first straight section is directed to the north; 
  * the southernmost (and if there are several, then the most western of among them) point of the track is the first point; 
  * the last point coincides with the first one (i.e., the start position); 
  * any pair of straight sections of the track has no shared points (except for the neighboring ones, they share exactly one point); 
  * no pair of points (except for the first and last one) is the same; 
  * no two adjacent straight sections are directed in the same direction or in opposite directions. 

Output

Print a single integer — the number of dangerous turns on the track.

Examples

Input

6
0 0
0 1
1 1
1 2
2 2
2 0
0 0


Output

1


Input

16
1 1
1 5
3 5
3 7
2 7
2 9
6 9
6 7
5 7
5 3
4 3
4 4
3 4
3 2
5 2
5 1
1 1


Output

6

Note

The first sample corresponds to the picture:

<image>

The picture shows that you can get in the water under unfortunate circumstances only at turn at the point (1, 1). Thus, the answer is 1.

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class BicycleRace {
	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		Point[] pts = new Point[n + 1];
		for (int i = 0; i <= n; ++i)
			pts[i] = new Point(sc.nextDouble(), sc.nextDouble());
		int cnt = 0;
		for (int i = 0; i + 1 < pts.length; ++i) 
			if(i<pts.length-2) {
				if(Point.ccw(pts[i], pts[i+1], pts[i+2]))
					++cnt;
			}
			else
				if(Point.ccw(pts[i], pts[i+1], pts[1])) 
					++cnt;
				
		System.out.println(cnt);
	}

	static class Point implements Comparable<Point> {

		static final double EPS = 1e-9;

		double x, y;

		Point(double a, double b) {
			x = a;
			y = b;
		}

		public String toString() {
			return x + " " + y;
		}

		public int compareTo(Point p) {
			if (Math.abs(x - p.x) > EPS)
				return x > p.x ? 1 : -1;
			if (Math.abs(y - p.y) > EPS)
				return y > p.y ? 1 : -1;
			return 0;
		}

		public double dist(Point p) {
			return Math.sqrt(sq(x - p.x) + sq(y - p.y));
		}

		static double sq(double x) {
			return x * x;
		}

		Point rotate(double angle) {
			double c = Math.cos(angle), s = Math.sin(angle);
			return new Point(x * c - y * s, x * s + y * c);
		}
		// for integer points and rotation by 90 (counterclockwise) : swap x and y,
		// negate x

		Point rotate(double theta, Point p) // rotate around p
		{
			Vector v = new Vector(p, new Point(0, 0));
			return translate(v).rotate(theta).translate(v.reverse());
		}

		Point translate(Vector v) {
			return new Point(x + v.x, y + v.y);
		}

		boolean between(Point p, Point q) {
			return x < Math.max(p.x, q.x) + EPS && x + EPS > Math.min(p.x, q.x) && y < Math.max(p.y, q.y) + EPS
					&& y + EPS > Math.min(p.y, q.y);
		}

		// returns true if it is on the line defined by a and b
		boolean onLine(Point a, Point b) {
			if (a.compareTo(b) == 0)
				return compareTo(a) == 0;
			return Math.abs(new Vector(a, b).cross(new Vector(a, this))) < EPS;
		}

		boolean onSegment(Point a, Point b) {
			if (a.compareTo(b) == 0)
				return compareTo(a) == 0;
			return onRay(a, b) && onRay(b, a);
		}

		// returns true if it is on the ray whose start point is a and passes through b
		boolean onRay(Point a, Point b) {
			if (a.compareTo(b) == 0)
				return compareTo(a) == 0;
			return new Vector(a, b).normalize().equals(new Vector(a, this).normalize()); // implement equals()
		}

		// returns true if it is on the left side of Line pq
		// add EPS to LHS if on-line points are accepted
		static boolean ccw(Point p, Point q, Point r) {
			return new Vector(p, q).cross(new Vector(p, r)) > 0;
		}

		static boolean collinear(Point p, Point q, Point r) {
			return Math.abs(new Vector(p, q).cross(new Vector(p, r))) < EPS;
		}

		static double angle(Point a, Point o, Point b) // angle AOB
		{
			Vector oa = new Vector(o, a), ob = new Vector(o, b);
			return Math.acos(oa.dot(ob) / Math.sqrt(oa.norm2() * ob.norm2()));
		}

		static double distToLine(Point p, Point a, Point b) // distance between point p and a line defined by points a,
															// b (a != b)
		{
			if (a.compareTo(b) == 0)
				return p.dist(a);
			// formula: c = a + u * ab
			Vector ap = new Vector(a, p), ab = new Vector(a, b);
			double u = ap.dot(ab) / ab.norm2();
			Point c = a.translate(ab.scale(u));
			return p.dist(c);
		}
		// Another way: find closest point and calculate the distance between it and p

		static double distToLineSegment(Point p, Point a, Point b) {
			Vector ap = new Vector(a, p), ab = new Vector(a, b);
			double u = ap.dot(ab) / ab.norm2();
			if (u < 0.0)
				return p.dist(a);
			if (u > 1.0)
				return p.dist(b);
			return distToLine(p, a, b);
		}
		// Another way: find closest point and calculate the distance between it and p
	}

	static class Vector {

		double x, y;

		Vector(double a, double b) {
			x = a;
			y = b;
		}

		Vector(Point a, Point b) {
			this(b.x - a.x, b.y - a.y);
		}

		Vector scale(double s) {
			return new Vector(x * s, y * s);
		} // s is a non-negative value

		double dot(Vector v) {
			return (x * v.x + y * v.y);
		}

		double cross(Vector v) {
			return x * v.y - y * v.x;
		}

		double norm2() {
			return x * x + y * y;
		}

		Vector reverse() {
			return new Vector(-x, -y);
		}

		Vector normalize() {
			double d = Math.sqrt(norm2());
			return scale(1 / d);
		}
	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		public long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}

		public double nextDouble() throws IOException {
			String x = next();
			StringBuilder sb = new StringBuilder("0");
			double res = 0, f = 1;
			boolean dec = false, neg = false;
			int start = 0;
			if (x.charAt(0) == '-') {
				neg = true;
				start++;
			}
			for (int i = start; i < x.length(); i++)
				if (x.charAt(i) == '.') {
					res = Long.parseLong(sb.toString());
					sb = new StringBuilder("0");
					dec = true;
				} else {
					sb.append(x.charAt(i));
					if (dec)
						f *= 10;
				}
			res += Long.parseLong(sb.toString()) / f;
			return res * (neg ? -1 : 1);
		}

		public boolean ready() throws IOException {
			return br.ready();
		}

	}
}