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())
ST = [list(map(int, input().split())) for _ in range(M)]
ST.sort()
Edge_out = [[] for _ in range(N+1)]
Edge_in = [[] for _ in range(N+1)]
for s, t in ST:
    Edge_out[s].append(t)
    Edge_in[t].append(s)

P_node = [0] * (N+1)  # 辿り着く確率
P_edge = [0] * (N+1)  # そこから出る辺を使う確率
P_node[1] = 1
for v, edge_out in enumerate(Edge_out[1:-1], 1):
    deg = len(edge_out)
    p = P_node[v] / deg
    P_edge[v] = p
    for u in edge_out:
        P_node[u] += p

# E_node_in = [0] * (N+1)  # ノードに到着するまでの辺の個数の期待値
#
# for v, edge_in in enumerate(Edge_in[2:], 2):
#     p_sum = 0
#     su = 0
#     for u in edge_in:
#         p = P_edge[u]
#         p_sum += p
#         su += p * E_node_in[u]
#     su /= p_sum
#     su += 1
#     E_node_in[v] = su

E_node_out = [0] * (N+1)  # ノードから出発したあと通る辺の個数の期待値

for v in range(N-1, 0, -1):
    edge_out = Edge_out[v]
    su = 0
    for u in edge_out:
        su += E_node_out[u]
    su /= len(edge_out)
    su += 1
    E_node_out[v] = su


# ans = e_all = E_node_out[1]
# for s, t in ST:
#     p_st = P_edge[s]
#     e_st = E_node_in[s] + E_node_out[t] + 1
#     print("e_st,p_st=",e_st,p_st)
#     a = (e_all - p_st*e_st) / (1 - p_st)
#     print("s,t,a=",s,t,a)
#     ans = min(ans, a)

ans = 0
for s, t in ST:
    p_s = P_node[s]
    #p_st = P_edge[s]
    deg = len(Edge_out[s])
    if deg==1:
        continue
    e = (E_node_out[s] - (1+E_node_out[t]) / deg) * (deg / (deg - 1))
    a = p_s * (e - E_node_out[s])
    #print("s,t,a=", s, t, a)
    ans = min(ans, a)
print(ans + E_node_out[1])