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:
n, m = map(int, input().split())
g = [[] for _ in range(n)]
ig = [[] for _ in range(n)]
z = []

for i in range(m):
    s, t = map(int, input().split())
    g[s-1].append(t-1)
    ig[t-1].append(s-1)
    z.append((s, t))

dp = [0]*n
for v in range(n-1)[::-1]:#DAGだから可能
    now = 0
    for u in g[v]:#DAGだからできる
        now += dp[u]
    now /= len(g[v])
    now += 1
    dp[v] = now

p = [0]*(n)
p[0] = 1
#削除したとき用：
for i in range(1, n):
    for _p in ig[i]:
        p[i]+=p[_p]/len(g[_p])

ma = 0
for s, t in z:#削除する矢印
    k = len(g[s-1])
    if k ==1 : continue
    a, b = dp[s-1], dp[t-1]
    c = k/(k-1)*(a-1-b/k) + 1
    ma = max(ma, (a-c)*p[s-1])
print(dp[0]-ma)