Problem:
Score : 600 points

Problem Statement

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}.

Sample Input 1

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

Sample Output 1

1.5000000000

If Aoki blocks the passage from Room 1 to Room 2, Takahashi will go along the path 1 → 3 → 4 with probability \frac{1}{2} and 1 → 4 with probability \frac{1}{2}. E = 1.5 here, and this is the minimum possible value of E.

Sample Input 2

3 2
1 2
2 3

Sample Output 2

2.0000000000

Blocking any one passage makes Takahashi unable to reach Room N, so Aoki cannot block a passage.

Sample Input 3

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

Sample Output 3

3.0133333333

Solution in PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)

def calc(N, adj, ignore, dp, prevs):
    s, t = ignore
    for i in range(prevs+1):
        dp[i] = 0.0
    for i in range(prevs, -1, -1):
        n = len(adj[i])
        if i == s:
            n -= 1
            if n == 0:
                return float("Inf")
        for nxt in adj[i]:
            if i == s and nxt == t:
                continue
            dp[i] += (dp[nxt] + 1) / n
    return dp[0]


def main():
    N, M = map(int, input().split())
    adj = [set() for _ in range(N)]
    x = [(-1, -1)]
    for _ in range(M):
        s, t = map(int, input().split())
        adj[s-1].add(t-1)
        x.append((s-1, t-1))
    x.sort(reverse=True)
    ans = float("Inf")
    dp = [0.0] * (N)
    dp[N-1] = 0.0
    prevs = N - 1
    for ignore in x:
        tmp = calc(N, adj, ignore, dp, prevs)
        ans = min(ans, tmp)
        prevs = ignore[0]
    print(ans)


if __name__ == "__main__":
    main()