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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
#define next Next
#define gc getchar
const int N=605;
int n,m,x[N],y[N],du[N],D[N],vis[N];
double ans,f[N];
vector<int>g[N];
/*char buf[1<<21],*p1=buf,*p2=buf;
inline int gc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}*/
inline int read()
{
    int ret=0,f=0;char c=gc();
    while(!isdigit(c)){if(c=='-')f=1;c=gc();}
    while(isdigit(c)){ret=ret*10+c-48;c=gc();}
    if(f)return -ret;return ret;
}
signed main()
{
	n=read();m=read();
	for(int i=1;i<=m;i++)x[i]=read(),y[i]=read();
	ans=1e9;
	for(int i=1;i<=m+1;i++)
	{
		for(int j=1;j<=n;j++)
		{
			vis[j]=0;du[j]=0;D[j]=0;f[j]=0.0;
			g[j].clear();
		}
		for(int j=1;j<=m;j++)
			if(i!=j)
			{
				du[x[j]]++;
				D[x[j]]++;
				g[y[j]].push_back(x[j]);
			}
		if(du[x[i]]==0)
		{
			for(int j=0;j<g[x[i]].size();j++)
			{
				int v=g[x[i]][j];
				vis[v]=1;
			}
		}
		queue<int>q;
		q.push(n);
		if(!du[x[i]])q.push(x[i]);
		while(!q.empty())
		{
			int u=q.front();q.pop();
			if(u==x[i]&&du[x[i]]==0)
			{
				for(int j=0;j<g[u].size();j++)
				{
					int v=g[u][j];
					D[v]--;
					if(!D[v])q.push(v);
				}
				continue;
			}
			if(vis[u])
			{
				f[u]=1.0*f[u]+1.0/du[u];
				f[u]=1.0*f[u]*du[u]/(du[u]-1);
			}
			for(int j=0;j<g[u].size();j++)
			{
				int v=g[u][j];
				D[v]--;
				f[v]+=1.0*(f[u]+1)/du[v];
				if(!D[v])q.push(v);
			}
		}
		if(f[1]>0.0)ans=min(ans,f[1]);
//		for(int j=1;j<=n;j++)printf("%.8lf\n",f[j]);
	}
	printf("%.8lf\n",ans);
	return 0;
}