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:
import sys
input = sys.stdin.readline
N, M = map(int, input().split())

g = [[] for _ in range(N+1)]
ig = [[] for _ in range(N+1)]
for _ in range(M):
    x, y = map(int, input().split())
    g[x].append(y)
    ig[y].append(x)

dp1 = [0] * (N+1)
dp2 = [0] * (N+1)
dp1[1] = 1
# dp1 : iまでの到達確率
# dp2 : iからゴールへの到達距離期待値

for i in range(1, N+1):
    for j in ig[i]:
        dp1[i] += dp1[j]/len(g[j])
df = 0
for i in range(N, 0, -1):
    t = 0
    for v in g[i]:
        t += (dp2[v]+1)
    if len(g[i]) > 0:
        dp2[i] = t/len(g[i])
    if len(g[i]) == 1:
        continue
    for v in g[i]:
        pv = (t-dp2[v]-1)/(len(g[i])-1)
        # pv : i→vへのpathを削除した時の到達距離期待値
        df = max(df, dp1[i]*(dp2[i]-pv))
        # dp1[i]*(dp2[i]-pv) : iからvを削除したときに1からvの到達距離期待値がどれだけ減るか
print(dp2[1]-df)