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>
int up(const void*a,const void*b){
    return*(int*)a<*(int*)b?-1:*(int*)a>*(int*)b?1:0;
}

void sort(int*a,int n){
    qsort(a,n,sizeof(int),up);
}
double min(double a,double b){
    return(a<b)?a:b;
}

double tansaku(int n,int m,int* a,int*b){
    int cango[n];
    double e[n+1];
    for(int i=1;i<n+1;i++)e[i]=0;
    int k;
    for(int i=n-1;i>0;i--){
        k=0;
        for(int j=0;j<m;j++){
            if(a[j]==i){
                cango[k]=b[j];
                k++;
            }
        }
        if(k==0)return 0;
        for(int j=0;j<k;j++){
            e[i]+=e[cango[j]];
        }
        e[i]/=k;
        e[i]++;
    }
    return e[1];
}

int main(void){
    int n,m;
    scanf("%d%d",&n,&m);
    int s[m],t[m],S[m],T[m];
    double EXP,EXP_tmp;
    for(int i=0;i<m;i++)scanf("%d%d",s+i,t+i);
    EXP=tansaku(n,m,s,t);
    for(int i=0;i<m;i++){
        int j=0,k=0;
        while(k<m){
            if(k!=i){
                S[j]=s[k];
                T[j]=t[k];
                j++;
            }
            k++;
        }
        EXP_tmp=tansaku(n,m-1,S,T);
        if(EXP_tmp>0)EXP=min(EXP,EXP_tmp);
    }
    printf("%f\n",EXP);
    return 0;
}