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 <string.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdlib.h>
#include <limits.h>
#include <math.h>
#include <float.h>
#include <assert.h>
typedef int64_t ll;
typedef uint64_t ull;
int acs(const void *a, const void *b){return *(int*)a - *(int*)b;} /* 1,2,3,4.. */
int des(const void *a, const void *b){return *(int*)b - *(int*)a;} /* 8,7,6,5.. */
#define min(a,b) (a < b ? a: b)
#define max(a,b) (a > b ? a: b)

#define MAXN (200001)
#define MOD (1000000007)


//グラフ
#define MAXV (MAXN)
int Gv;
typedef struct edge
{
    int to;
    ll cost;
    struct edge* next;
}edge;

edge* Ge[MAXV]; //vから出てる有向辺
int Gd[MAXV]; //vの次数

void Gadd(int v, int to, ll cost)
{
    edge* e = malloc(sizeof(edge));
    e->to = to;
    e->cost = cost;
    e->next = Ge[v];
    Ge[v] = e;
    Gd[v]++;
}

int n;
double sc[MAXV];

int main(void)
{
    ll m;
    scanf("%d %ld",&n,&m);
    for(int i=0;i<m;i++) 
    {
        int s,t;
        scanf("%d %d",&s,&t);
        s--;t--;
        Gadd(s,t,0);
    }

    double ans = DBL_MAX;
    for(int i=0;i<n-1;i++)
    {
        for(int j=0;j<n;j++) sc[j] = -1.0;
        sc[n-1] = 0.0;
        for(int j=n-2;j>=0;j--)
        {
            double mx = 0.0;
            double ns = 0.0;
            int c=0;
            edge* e = Ge[j];
            for(;e!=NULL;e=e->next)
            {
                if(sc[e->to]==-1.0) continue;
                mx = max(mx,sc[e->to]+1.0);
                ns += sc[e->to]+1.0;
                c++;
            }
            if(ns==0.0) continue;
            if(i==j && c>=2)
            {
                sc[j] = (ns-mx)/(c-1);
            }else
            {
                sc[j] = ns/c;
            }
        }
        ans = min(ans,sc[0]);
    }

    printf("%lf\n", ans);
    return 0;
}