Problem:
Score : 600 points

Problem Statement

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}.

Sample Input 1

4 6
1 4
2 3
1 3
1 2
3 4
2 4

Sample Output 1

1.5000000000

If Aoki blocks the passage from Room 1 to Room 2, Takahashi will go along the path 1 → 3 → 4 with probability \frac{1}{2} and 1 → 4 with probability \frac{1}{2}. E = 1.5 here, and this is the minimum possible value of E.

Sample Input 2

3 2
1 2
2 3

Sample Output 2

2.0000000000

Blocking any one passage makes Takahashi unable to reach Room N, so Aoki cannot block a passage.

Sample Input 3

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

Sample Output 3

3.0133333333

Solution in OCaml (4.02.3) (judge verdict: Accepted)

let scan_int () = Scanf.scanf " %d" (fun x -> x)
let print_float n = Printf.printf "%.12f\n" n

(* 辺の入力 *)
let input_edge edges m =
  for i = 0 to m-1 do
    let s = scan_int() and t = scan_int() in
    edges.(s-1) <- (t-1) :: edges.(s-1) 
  done

(* 頂点vからの期待値の計算 *)
let calc_expected_value n edges =
  let expected = Array.make n 0.0 in
  let prop expect p u v = expect.(u) <- expect.(u) +. (expect.(v) +. 1.0) *. p in
  let rec rev_calc curr edges ret =
    if curr < 0 then ret
    else
      let p = 1.0 /. (float (List.length edges.(curr))) in
      let _ = List.iter (prop ret p curr) edges.(curr) in
      rev_calc (curr-1) edges ret in
  rev_calc (n-1) edges expected

(* 頂点vを通る確率の計算 *)
let calc_probability n edges =
  let probability = Array.make n 0.0 in
  let _ = probability.(0) <- 1.0 in
  let prop prob p u v = prob.(v) <- prob.(v) +. prob.(u) *. p in
  let rec calc curr edges ret =
    if curr = n then ret
    else
      let p = 1.0 /. (float (List.length edges.(curr))) in
      let _ = List.iter (prop ret p curr) edges.(curr) in
      calc (curr+1) edges ret in
  calc 0 edges probability


let solve n edges expected probability =
  let ans = ref expected.(0) in
  let p = expected.(0) in
  let _ =
    for u = 0 to n-1 do
      let l = List.length edges.(u) in
      if l <= 1 then ()
      else
        let add_prob = 1.0 /. (float (l-1)) -. 1.0 /. (float l) in
        let fold_func p v = p +. add_prob *. expected.(v) *. probability.(u) in
        let ref_value = List.fold_left fold_func p edges.(u) in
        let sub_prob = 1.0 /. (float (l-1)) in
        let calc_func v =
          let result = ref_value -. sub_prob *. expected.(v) *. probability.(u) in
          ans := min !ans result in
        List.iter calc_func edges.(u)
    done in
  print_float !ans


let () =
  let n = scan_int() and m = scan_int() in
  let edges = Array.make n [] in
  let _ = input_edge edges m in
  let expected = calc_expected_value n edges in
  let probability = calc_probability n edges in
  solve n edges expected probability