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 C (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include<stdlib.h>



typedef struct edge{
  int name;
  struct edge *next;
} Edge;

typedef struct node{
  double e;
  int kazu;
  struct edge *to;
} Node;


void insert(Node *list[],int from,int to){
  Node *ntmp = list[from-1];
  if(ntmp->to == NULL){
    ntmp->kazu=1;
    Edge *new = malloc(sizeof(struct edge));
    new->name = to;
    new->next = NULL;
    ntmp->to = new;}
  else{
    ntmp->kazu++;
    Edge *tmp = list[from-1]->to;
    while(tmp->next!=NULL){
      tmp=tmp->next;}
    Edge *new = malloc(sizeof(struct edge));
    new->name = to;
    new->next = NULL;
    tmp->next = new;}
}

int main(){
  int n,m;
  scanf("%d %d\n",&n,&m);

  //list of nodes
  Node *list[n];
  for(int i=0;i<n;i++){
    list[i] = malloc(sizeof(Node));
    list[i]->to = NULL;}
  //printf("nm:%d %d\n",m,n);

  //insert nodes
  for(int i=0;i<m;i++){
    int from,to;
    scanf("%d%d",&from,&to);
    //printf("input:%d %d\n",from,to);
    insert(list,from,to);}
  //printf("insert finished\n");


  double min=m;

  //calculate
  list[n-1]->e=0;
  for(int l=0;l<n-1;l++){
  for(int i=n-1;i>0;i--){
    double max=0;
    int j=i-1;
    Node *ntmp=list[j];
    Edge *etmp=ntmp->to;
    ntmp->e = 0;
    while(etmp!=NULL){
      ntmp->e += (list[etmp->name-1]->e+1);
      if(max < list[etmp->name-1]->e){
	max= list[etmp->name-1]->e;}
      etmp=etmp->next;}
    if(l==j && ntmp->kazu >1){
      ntmp->e = 1.0* (ntmp->e-max-1)/(ntmp->kazu-1);}
    else{
      ntmp->e = 1.0* ntmp->e/ntmp->kazu;}}
    if(min > list[0]->e){
      min=list[0]->e;}}
  //printf("calc finished\n");

       /* 
  //calulate 1
  if(list[0]->kazu > 1){
    double max=0;
    Edge *tmp=list[0]->to;
    list[0]->e=0;
    while(tmp!=NULL){
      list[0]->e += list[tmp->name-1]->e+1;
      if(max < list[tmp->name-1]->e){
	max= list[tmp->name-1]->e;}
      tmp=tmp->next;}
    //printf("%f\n",max);
    list[0]->e = 1.0*(list[0]->e - (max+1))/(list[0]->kazu-1);}
    //printf("calc1 finished\n");*/

  /*
  for(int i=n-1;i>=0;i--){
    printf("%f\n",list[i]->e);}*/
  printf("%f\n",min);

  return 0;
}