Problem: Flatland has recently introduced a new type of an eye check for the driver's licence. The check goes like that: there is a plane with mannequins standing on it. You should tell the value of the minimum angle with the vertex at the origin of coordinates and with all mannequins standing inside or on the boarder of this angle. 

As you spend lots of time "glued to the screen", your vision is impaired. So you have to write a program that will pass the check for you.

Input

The first line contains a single integer n (1 ≤ n ≤ 105) — the number of mannequins.

Next n lines contain two space-separated integers each: xi, yi (|xi|, |yi| ≤ 1000) — the coordinates of the i-th mannequin. It is guaranteed that the origin of the coordinates has no mannequin. It is guaranteed that no two mannequins are located in the same point on the plane.

Output

Print a single real number — the value of the sought angle in degrees. The answer will be considered valid if the relative or absolute error doesn't exceed 10 - 6. 

Examples

Input

2
2 0
0 2


Output

90.0000000000


Input

3
2 0
0 2
-2 2


Output

135.0000000000


Input

4
2 0
0 2
-2 0
0 -2


Output

270.0000000000


Input

2
2 1
1 2


Output

36.8698976458

Note

Solution for the first sample test is shown below: 

<image>

Solution for the second sample test is shown below: 

<image>

Solution for the third sample test is shown below: 

<image>

Solution for the fourth sample test is shown below: 

<image>

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

import static java.lang.Math.*;
import static java.util.Arrays.*;

public class ladders257c {
    static BufferedReader __in;
    static PrintWriter __out;
    static StringTokenizer input;

    public static void main(String[] args) throws IOException {
        __in = new BufferedReader(new InputStreamReader(System.in));
        __out = new PrintWriter(new OutputStreamWriter(System.out));

        int n = readInt();
        double[] a = new double[n];
        for(int i = 0; i < n; ++i) {
            read();
            int x = nextInt(), y = nextInt();
            a[i] = atan2(y, x);
            if(a[i] < 0) {
                a[i] += PI * 2;
            }
        }
        if(n == 1) {
            println(0);
            close();
            return;
        }
        sort(a);
        double max = 0;
        for(int i = 0; i < n - 1; ++i) {
            double diff = a[i + 1] - a[i];
            if(diff > max) {
                max = diff;
            }
        }
        double diff = a[0] + PI * 2 - a[a.length - 1];
        if(diff > max) {
            max = diff;
        }
        println(toDegrees(2 * PI - max));

        close();
    }

    static final int IBIG = 1000000007;
    static final int IMAX = 2147483647;
    static final long LMAX = 9223372036854775807L;
    static void read() throws IOException {input = new StringTokenizer(__in.readLine());}
    static String readLine() throws IOException {return __in.readLine();}
    static int nextInt() {return Integer.parseInt(input.nextToken());}
    static int readInt() throws IOException {return Integer.parseInt(__in.readLine());}
    static void print(String s) {__out.print(s);}
    static void println(String s) {__out.println(s);}
    static void print(Object o) {__out.print(o);}
    static void println(Object o) {__out.println(o);}
    static void printy() {__out.println("Yes");}
    static void printY() {__out.println("YES");}
    static void printn() {__out.println("No");}
    static void printN() {__out.println("NO");}
    static void printyn(boolean b) {__out.println(b ? "Yes" : "No");}
    static void printYN(boolean b) {__out.println(b ? "YES" : "NO");}
    static void println(int... a) {for(int i = 0, len = a.length - 1; i < len; __out.print(a[i]), __out.print(' '), i++); __out.println(a[a.length - 1]);}
    static void println() {__out.println();}
    static void close() {__out.close();}
}