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 PYTHON3):

Solution:
def f_fork_in_the_road():
    # 参考: https://www.onakasuitacity.com/abc144-f/
    N, M = [int(i) for i in input().split()]
    Nodes = [[int(i) for i in input().split()] for j in range(M)]

    edges = [[] for _ in range(N)]
    for s, t in Nodes:
        edges[s - 1].append(t - 1)

    # editorial の DP テーブルの確率部分
    probability = [0] * N
    probability[0] = 1.0
    for v in range(N):
        for next_vertex in edges[v]:
            probability[next_vertex] += probability[v] / len(edges[v])

    # 期待値
    expected_value = [0] * N
    for v in range(N - 2, -1, -1):
        for next_vertex in edges[v]:
            expected_value[v] += expected_value[next_vertex]
        expected_value[v] = 1 + expected_value[v] / len(edges[v])

    ans = expected_value[0]
    for v in range(N):
        if len(edges[v]) <= 1:
            continue
        # 各頂点 v に対して、v から出る辺 vw のうち
        # w での期待値が最大のものの分を差分から引く。残りは足す
        v_max_expected = max(((expected_value[w], w) for w in edges[v]))[1]
        expected_diff = 0.0
        outdegree = len(edges[v])
        for next_vertex in edges[v]:
            if next_vertex == v_max_expected:
                expected_diff -= expected_value[next_vertex] / outdegree
            else:
                expected_diff += expected_value[next_vertex] / (outdegree * (outdegree - 1))
        ans = min(ans, expected_value[0] + probability[v] * expected_diff)
    return round(ans, 10)

print(f_fork_in_the_road())