Problem: You are given a convex polygon P with n distinct vertices p1, p2, ..., pn. Vertex pi has coordinates (xi, yi) in the 2D plane. These vertices are listed in clockwise order.

You can choose a real number D and move each vertex of the polygon a distance of at most D from their original positions.

Find the maximum value of D such that no matter how you move the vertices, the polygon does not intersect itself and stays convex.

Input

The first line has one integer n (4 ≤ n ≤ 1 000) — the number of vertices.

The next n lines contain the coordinates of the vertices. Line i contains two integers xi and yi ( - 109 ≤ xi, yi ≤ 109) — the coordinates of the i-th vertex. These points are guaranteed to be given in clockwise order, and will form a strictly convex polygon (in particular, no three consecutive points lie on the same straight line).

Output

Print one real number D, which is the maximum real number such that no matter how you move the vertices, the polygon stays convex.

Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.

Namely, let's assume that your answer is a and the answer of the jury is b. The checker program will consider your answer correct if <image>.

Examples

Input

4
0 0
0 1
1 1
1 0


Output

0.3535533906


Input

6
5 0
10 0
12 -4
10 -8
5 -8
3 -4


Output

1.0000000000

Note

Here is a picture of the first sample

<image>

Here is an example of making the polygon non-convex.

<image>

This is not an optimal solution, since the maximum distance we moved one point is  ≈ 0.4242640687, whereas we can make it non-convex by only moving each point a distance of at most  ≈ 0.3535533906.

Solution (in JAVA):

Solution:
import java.util.*;
import java.awt.geom.Line2D;
import java.io.*;
public class VolatileKite {
    
    
    
    /************************ SOLUTION STARTS HERE ************************/
    static double dot(double[] A, double[] B, double[] C){
        double[] AB = new double[2];
        double[] BC = new double[2];
        AB[0] = B[0]-A[0];
        AB[1] = B[1]-A[1];
        BC[0] = C[0]-B[0];
        BC[1] = C[1]-B[1];
        double dot = AB[0] * BC[0] + AB[1] * BC[1];
        return dot;
    }
    //Compute the cross product AB x AC
    static double cross(double[] A, double[] B, double[] C){
        double[] AB = new double[2];
        double[] AC = new double[2];
        AB[0] = B[0]-A[0];
        AB[1] = B[1]-A[1];
        AC[0] = C[0]-A[0];
        AC[1] = C[1]-A[1];
        double cross = AB[0] * AC[1] - AB[1] * AC[0];
        return cross;
    }
    //Compute the distance from A to B
    static double distance(double[] A, double[] B){
        double d1 = A[0] - B[0];
        double d2 = A[1] - B[1];
        return Math.sqrt(d1*d1+d2*d2);
    }
    //Compute the distance from AB to C
    //if isSegment is true, AB is a segment, not a line.
    static double linePointDist(double[] A, double[] B, double[] C, boolean isSegment){
        double dist = cross(A,B,C) / distance(A,B);
        if(isSegment){
            double dot1 = dot(A,B,C);
            if(dot1 > 0)return distance(B,C);
            double dot2 = dot(B,A,C);
            if(dot2 > 0)return distance(A,C);
        }
        return Math.abs(dist);
    }
  

    private static void solve() {
        
        
        int n = nextInt();
        double p[][] = new double[n][2];
        for(int i = 0; i < n; i++) {
            p[i][0] = nextDouble();
            p[i][1] = nextDouble();
        }
        
        double min = Double.MAX_VALUE;
        for(int i = 0; i < n; i++) {
            int prev = (i - 1 + n) % n;
            int next = (i + 1) % n;
            min = Math.min(min , linePointDist(p[prev], p[next], p[i], true));
//            min = Math.min(Line2D.Double.ptSegDist(p[prev][0], p[prev][1], p[next][0], p[next][1], p[i][0], p[i][1]), min);
        }
        
        println(min / 2.0);
    }
    
    
    
    /************************ SOLUTION ENDS HERE ************************/
    
    
    
    
    
    /************************ TEMPLATE STARTS HERE **********************/
    
    public static void main(String[] args) throws IOException {
        reader = new BufferedReader(new InputStreamReader(System.in));
        writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)), false);
        st     = null;
        solve();
        reader.close();
        writer.close();
    }
    
    static BufferedReader reader;
    static PrintWriter    writer;
    static StringTokenizer st;
    
    static String next()
    {while(st == null || !st.hasMoreTokens()){try{String line = reader.readLine();if(line == null){return null;}            
    st = new StringTokenizer(line);}catch (Exception e){throw new RuntimeException();}}return st.nextToken();}
    static String nextLine()  {String s=null;try{s=reader.readLine();}catch(IOException e){e.printStackTrace();}return s;}             
    static int    nextInt()   {return Integer.parseInt(next());}
    static long   nextLong()  {return Long.parseLong(next());}     
    static double nextDouble(){return Double.parseDouble(next());}
    static char   nextChar()  {return next().charAt(0);}
    static int[]  nextIntArray(int n)         {int[] a= new int[n];   int i=0;while(i<n){a[i++]=nextInt();}  return a;}
    static long[] nextLongArray(int n)        {long[]a= new long[n];  int i=0;while(i<n){a[i++]=nextLong();} return a;}    
    static int[]  nextIntArrayOneBased(int n) {int[] a= new int[n+1]; int i=1;while(i<=n){a[i++]=nextInt();} return a;}            
    static long[] nextLongArrayOneBased(int n){long[]a= new long[n+1];int i=1;while(i<=n){a[i++]=nextLong();}return a;}            
    static void   print(Object o)  { writer.print(o);  }
    static void   println(Object o){ writer.println(o);}
    
    /************************ TEMPLATE ENDS HERE ************************/
    
}