Problem: There is a cave consisting of N rooms and M one-directional passages. The rooms are numbered 1 through N.

Takahashi is now in Room 1, and Room N has the exit. The i-th passage connects Room s_i and Room t_i (s_i < t_i) and can only be traversed in the direction from Room s_i to Room t_i. It is known that, for each room except Room N, there is at least one passage going from that room.

Takahashi will escape from the cave. Each time he reaches a room (assume that he has reached Room 1 at the beginning), he will choose a passage uniformly at random from the ones going from that room and take that passage.

Aoki, a friend of Takahashi's, can block one of the passages (or do nothing) before Takahashi leaves Room 1. However, it is not allowed to block a passage so that Takahashi is potentially unable to reach Room N.

Let E be the expected number of passages Takahashi takes before he reaches Room N. Find the value of E when Aoki makes a choice that minimizes E.

Constraints

* 2 \leq N \leq 600
* N-1 \leq M \leq \frac{N(N-1)}{2}
* s_i < t_i
* If i != j, (s_i, t_i) \neq (s_j, t_j). (Added 21:23 JST)
* For every v = 1, 2, ..., N-1, there exists i such that v = s_i.

Input

Input is given from Standard Input in the following format:


N M
s_1 t_1
:
s_M t_M


Output

Print the value of E when Aoki makes a choice that minimizes E. Your output will be judged as correct when the absolute or relative error from the judge's output is at most 10^{-6}.

Examples

Input

4 6
1 4
2 3
1 3
1 2
3 4
2 4


Output

1.5000000000


Input

3 2
1 2
2 3


Output

2.0000000000


Input

10 33
3 7
5 10
8 9
1 10
4 6
2 5
1 7
6 10
1 4
1 3
8 10
1 5
2 6
6 9
5 6
5 8
3 6
4 8
2 7
2 9
6 7
1 2
5 9
6 8
9 10
3 9
7 8
4 5
2 10
5 7
3 5
4 7
4 9


Output

3.0133333333

Solution (in JAVA):

Solution:
import java.util.*;

public class Main {

    public static void main(String[] args){

        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        boolean[][] g = new boolean[n][n];
        int[] fo = new int[n];
        double[] p = new double[n];
        double[] l = new double[n];
        for(int i=0; i<m; i++) {
            int s = sc.nextInt() - 1;
            int t = sc.nextInt() - 1;
            g[s][t] = true;
            fo[s]++;
        }
        p[0] = 1d;
        for(int i=0; i<n; i++) {
            for(int j=i+1; j<n; j++) {
                if(g[i][j]) {
                    p[j] += p[i] / fo[i];
                }
            }
        }
        for(int i=n-2; i>=0; i--) {
            l[i] = 1;
            for(int j=i+1; j<n; j++) {
                if(g[i][j]) {
                    l[i] += l[j] / fo[i];
                }
            }
        }

        double res = l[0];

        for(int i=0; i<n; i++) {
            if(fo[i] < 2) continue;
            double max = 0;
            int maxj = 0;
            for(int j=i+1; j<n; j++) {
                if(g[i][j]) {
                    max = Math.max(max, l[j]);
                    if(l[j] == max) {
                        maxj = j;
                    }
                }
            }
            double l1 = 1;
            for(int j=i+1; j<n; j++) {
                if(g[i][j] && j != maxj) {
                    l1 += l[j] / (fo[i] - 1);
                }
            }
            double d = l[i] - l1;
            double l0 = l[0] - d * p[i];
            res = Math.min(res, l0);
        }

        System.out.println(res);
    }


}