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
EPS = 10**(-8)
INF = 10**16

N, M = map(int, input().split())
graph = [set() for _ in range(N)]
for _ in range(M):
    a, b = map(int, input().split())
    graph[a-1].add(b-1)

dp = [0]*N
dps = [0]*N
dpl = [0]*N
for n in reversed(range(N-1)):
    for p in graph[n]:
        if dp[p] != -1:
            dps[n] += dp[p]
            dpl[n] += 1
    if dpl[n] > 0:
        dp[n] = dps[n]/dpl[n] + 1
    else:
        dp[n] = -1

Weight = [0]*N
Weight[0] = 1
for n in range(N):
    for p in graph[n]:
        Weight[p] += Weight[n]/dpl[n]

def solve(n):
    dpt = dp[:]
    dpts = [0]*(n+1) + dps[n+1:]
    dptl = [0]*(n+1) + dpl[n+1:]
    dpt[n] = -1
    for p in reversed(range(n)):
        for q in graph[p]:
            if dpt[q] != -1:
                dpts[p] += dpt[q]
                dptl[p] += 1
        if dptl[p] > 0:
            dpt[p] = dpts[p]/dptl[p] + 1
        else:
            dpt[p] = -1
    if dpt[0] == -1: return INF
    return dpt[0]

ans = dp[0]
for n1 in range(N):
    if len(graph[n1]) == 1:
        continue
        #ans = min(ans, solve(n1))
    else:
        for n2 in graph[n1]:
            newE = (dps[n1]-dp[n2])/(dpl[n1]-1) + 1
            ans = min(ans, dp[0] - Weight[n1]*dp[n1] + Weight[n1]*newE)

print(ans)