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:
#!usr/bin/env python3
from collections import defaultdict,deque
from heapq import heappush, heappop
import sys
import math
import bisect
import random
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def LS():return [list(x) for x in sys.stdin.readline().split()]
def S():
    res = list(sys.stdin.readline())
    if res[-1] == "\n":
        return res[:-1]
    return res
def IR(n):
    return [I() for i in range(n)]
def LIR(n):
    return [LI() for i in range(n)]
def SR(n):
    return [S() for i in range(n)]
def LSR(n):
    return [LS() for i in range(n)]

sys.setrecursionlimit(1000000)
mod = 1000000007

def solve():
    n,m = LI()
    v = [[] for i in range(n)]
    v_ = [[] for i in range(n)]
    for i in range(m):
        a,b = LI()
        a -= 1
        b -= 1
        v[a].append(b)
        v_[b].append(a)
    dp = [0]*n
    for y in range(1,n)[::-1]:
        v_[y].sort()
        d = dp[y]
        for x in v_[y]:
            dp[x] += (d+1)/len(v[x])
    ans = dp[0]
    p = [0]*n
    p[0] = 1
    for x in range(n-1):
        d = p[x]/len(v[x])
        for y in v[x]:
            p[y] += d
    for a in range(n-1):
        if len(v[a]) == 1:
            continue
        m = max([dp[b] for b in v[a]])
        for b in v[a]:
            if dp[b] == m:
                break
        l = dp[0]
        k = dp[a]
        dp[a] = 0
        le = len(v[a])-1
        for y in v[a]:
            if y == b:
                continue
            dp[a] += (dp[y]+1)/le
        d = l+p[a]*(dp[a]-k)
        dp[0] = l
        if 0 < d < ans:
            ans = d
    print(ans)
    return

#Solve
if __name__ == "__main__":
    solve()