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 {

    static Scanner sc = new Scanner(System.in);

    public static void main(String[] args) {
        int N = sc.nextInt();
        int M = sc.nextInt();
        ArrayList<ArrayList<Integer>> g = new ArrayList<>();
        for (int i = 0; i < N; i++) {
            g.add(new ArrayList<>());
        }
        for (int i = 0; i < M; i++) {
            int S = sc.nextInt() - 1;
            int T = sc.nextInt() - 1;
            g.get(S).add(T);
        }
        double ans = 1e10;
        double[] dp = new double[N];
        for (int i = 0; i < N; i++) {
            Arrays.fill(dp, 0.0);
            for (int j = N - 2; j >= 0; j--) {
                double sum = 0;
                double max = 0;
                for (int n : g.get(j)) {
                    sum += dp[n];
                    max = Math.max(max, dp[n]);
                }
                dp[j] = sum / g.get(j).size() + 1;
                if (j == i && g.get(j).size() > 1) {
                    dp[j] = (sum - max) / (g.get(j).size() - 1) + 1;
                }
            }
            ans = Math.min(ans, dp[0]);
        }
        System.out.printf("%.9f\n", ans);
    }
}