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
sys.setrecursionlimit(2147483647)
INF=float("inf")
MOD=10**9+7
input=lambda :sys.stdin.readline().rstrip()
def resolve():
    n,m=map(int,input().split())
    E=[[] for _ in range(n)]
    for _ in range(m):
        s,t=map(int,input().split())
        s-=1; t-=1;
        E[s].append(t)
    dp=[0]*n
    for v in range(n-2,-1,-1):
        res=0
        for nv in E[v]: res+=dp[nv]
        dp[v]=1+res/len(E[v])
    ans=dp[0] # 辺を消さなかった時のscore

    # 各 v に対し、削除する辺 v→u は、dp[u]が最大のもの
    for v in range(n-1):
        if(len(E[v])==1): continue
        u=max(*((dp[nv],nv) for nv in E[v]))[1]
        # vu を削除したときの期待値DPを行う
        dp2=[0]*n
        for x in range(n-2,-1,-1):
            res=0
            for nx in E[x]:
                if(x==v and nx==u): continue
                res+=dp2[nx]
            if(v!=x):
                dp2[x]=1+res/len(E[x])
            else:
                dp2[x]=1+res/(len(E[x])-1)
        ans=min(ans,dp2[0])
    print(ans)
resolve()