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:
def solve():
    import sys
    input = sys.stdin.readline

    N, M = map(int, input().split())
    adjL = [[] for _ in range(N)]
    outdegs = [0] * N
    for _ in range(M):
        s, t = map(int, input().split())
        s, t = s-1, t-1
        adjL[s].append(t)
        outdegs[s] += 1

    probs = [0] * N
    probs[0] = 1
    for v in range(N):
        for v2 in adjL[v]:
            probs[v2] += probs[v] / outdegs[v]

    dp = [0] * N
    for v in reversed(range(N-1)):
        dp[v] = sum([dp[v2] for v2 in adjL[v]]) / outdegs[v] + 1

    ans = dp[0]
    for vRem in range(N):
        if outdegs[vRem] == 1: continue
        sumDP, maxDP = 0, -1
        for v2 in adjL[vRem]:
            sumDP += dp[v2]
            if dp[v2] > maxDP:
                maxDP = dp[v2]
        dp2 = (sumDP-maxDP) / (outdegs[vRem]-1) + 1
        decre = dp[vRem] - dp2
        ans2 = dp[0] - decre * probs[vRem]
        ans = min(ans, ans2)

    print(ans)


solve()