Problem: The scientists have recently discovered wormholes — objects in space that allow to travel very long distances between galaxies and star systems. 

The scientists know that there are n galaxies within reach. You are in the galaxy number 1 and you need to get to the galaxy number n. To get from galaxy i to galaxy j, you need to fly onto a wormhole (i, j) and in exactly one galaxy day you will find yourself in galaxy j. 

Unfortunately, the required wormhole is not always available. Every galaxy day they disappear and appear at random. However, the state of wormholes does not change within one galaxy day. A wormhole from galaxy i to galaxy j exists during each galaxy day taken separately with probability pij. You can always find out what wormholes exist at the given moment. At each moment you can either travel to another galaxy through one of wormholes that exist at this moment or you can simply wait for one galaxy day to see which wormholes will lead from your current position at the next day.

Your task is to find the expected value of time needed to travel from galaxy 1 to galaxy n, if you act in the optimal way. It is guaranteed that this expected value exists.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 1000) — the number of galaxies within reach.

Then follows a matrix of n rows and n columns. Each element pij represents the probability that there is a wormhole from galaxy i to galaxy j. All the probabilities are given in percents and are integers. It is guaranteed that all the elements on the main diagonal are equal to 100.

Output

Print a single real value — the expected value of the time needed to travel from galaxy 1 to galaxy n if one acts in an optimal way. 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

3
100 50 50
0 100 80
0 0 100


Output

1.750000000000000


Input

2
100 30
40 100


Output

3.333333333333333

Note

In the second sample the wormhole from galaxy 1 to galaxy 2 appears every day with probability equal to 0.3. The expected value of days one needs to wait before this event occurs is <image>.

Solution (in JAVA):

Solution:
//package codeforces.cfr335div1;

import java.io.*;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.util.function.Function;

/**
 * Created by raggzy on 4/3/2016.
 */
public class E {

    private static class Reader implements Closeable {
        private BufferedReader br;
        private StringTokenizer st;

        public Reader(InputStream is) {
            br = new BufferedReader(new InputStreamReader(is));
        }

        public void readLine() {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }

        public String next() {
            if (st == null || !st.hasMoreTokens()) {
                readLine();
            }
            return st.nextToken();
        }

        public <T> T next(Function<String, T> parser) {
            return parser.apply(next());
        }

        public int nextInt() {
            return next(Integer::valueOf);
        }

        public long nextLong() {
            return next(Long::valueOf);
        }

        public double nexDouble() {
            return next(Double::valueOf);
        }

        @Override
        public void close() throws IOException {
            br.close();
        }
    }


    public static final double EPS = 1e-8;

    public static void main(String[] args) {
        Reader in = new Reader(System.in);
        int n = in.nextInt();
        double[][] p = new double[n][n];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                p[i][j] = 1.0 * in.nextInt() / 100;
            }
        }
        double[] f = new double[n];
        Arrays.fill(f, -1);
        double[] prob = new double[n];
        Arrays.fill(prob, 1.0);
        double[] sumAchieved = new double[n];
        int minI = n - 1;
        double minRes = 0;

        while (minI >= 0) {
            f[minI] = minRes;
            int nextMin = -1;
            double nextRes = -1;
            for (int i = 0; i < n; i++) {
                if (f[i] >= 0) continue;
                sumAchieved[i] += prob[i] * p[i][minI] * (1 + f[minI]);
                prob[i] *= (1 - p[i][minI]);
                if (prob[i] < 1) {
                    double res = (prob[i] + sumAchieved[i]) / (1. - prob[i]);
                    if (nextMin == -1 || res < nextRes) {
                        nextMin = i;
                        nextRes = res;
                    }
                }
            }
            minI = nextMin;
            minRes = nextRes;
        }
        System.out.println(f[0]);
    }
}