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())
info = [list(map(int, input().split())) for i in range(m)]

graph = [[] for i in range(n)]
for i in range(m):
    a, b = info[i]
    a -= 1
    b -= 1
    graph[a].append(b)

def dfs(pos):
    if dp[pos] != -1:
        return dp[pos]
    p = len(graph[pos])
    dp[pos] = 0
    if p == 0:
        return dp[pos]
    for next_pos in graph[pos]:
        dp[pos] += dfs(next_pos) / p
    dp[pos] += 1
    return dp[pos]

dp = [-1]*n
dfs(0)
ans = dp[0]

def dfs_1(pos):
    if dq[pos] != -1:
        return dq[pos]
    p = len(graph[pos])
    if pos == i:
        p -= 1
    dq[pos] = 0
    if p == 0:
        return dq[pos]
    for next_pos in graph[pos]:
        if pos == i and next_pos == ind:
            continue
        dq[pos] += dfs_1(next_pos) / p
    dq[pos] += 1
    return dq[pos]

for i in range(n):
    num = 0
    ind = 0
    if len(graph[i]) <= 1:
        continue
    for j in graph[i]:
        if num < dp[j]:
            num = dp[j]
            ind = j
    dq = [-1]*n
    dfs_1(0)
    ans = min(dq[0], ans)
print(ans)