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:
from collections import defaultdict


def main():
    N, M = list(map(int, input().split()))
    adj_from = defaultdict(list)
    adj_to = defaultdict(list)
    for _ in range(M):
        s, t = list(map(lambda x: int(x) - 1, input().split()))
        adj_from[t].append(s)
        adj_to[s].append(t)
    # prob[s][t]: 邪魔がない場合に枝s->tを通る確率
    prob = [[0] * N for _ in range(N)]
    for t in adj_to[0]:
        prob[0][t] = 1 / len(adj_to[0])
    for s in range(1, N - 1):
        # node 's' に入ってくる確率を均等に分配
        prob_sum = sum([prob[n][s] for n in adj_from[s]])
        for t in adj_to[s]:
            prob[s][t] += prob_sum / len(adj_to[s])
    # expect[s][t]: 邪魔がなく、かつ枝s->tからカウントを始めた場合に通る枝本数の期待値
    expect = [[0] * N for _ in range(N)]
    for s in adj_from[N - 1]:
        expect[s][N - 1] = 1
    for t in range(N - 2, 0, -1):
        # node 't' から出ていく時の本数を、node 't' に入ってくるところに反映
        e = sum([expect[t][n] for n in adj_to[t]]) / len(adj_to[t])
        for s in adj_from[t]:
            expect[s][t] += e + 1
    # 邪魔なしの場合の枝本数の期待値
    expect_all = sum([p * e for p, e in zip(prob[0], expect[0])])
    # 各edgeをふさいだ場合の枝本数の変化量の最小値を求める
    diff_min = 0  # ふさげない場合 / ふさぐと不利になる場合はふさがないので変化量0
    for s in range(N - 1):
        if len(adj_to[s]) <= 1:
            continue
        s_prob = 1 if s == 0 else sum([prob[n][s] for n in adj_from[s]])
        # expect[s][t] が最も大きいtを選んでふさぐのが良い
        expect_max, target = 0, None
        for t in adj_to[s]:
            if expect[s][t] > expect_max:
                expect_max = expect[s][t]
                target = t
        # 枝s->targetをふさいだ時の枝の期待値の変化量
        d = - prob[s][target] * expect[s][target]
        for t in adj_to[s]:
            if t == target:
                continue
            d += ((s_prob / (len(adj_to[s]) - 1)) - prob[s][t]) * expect[s][t]
        diff_min = min(diff_min, d)
    # 答え
    print(expect_all + diff_min)


if __name__ == '__main__':
    main()