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: Time Limit Exceeded)

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int INF=0x3f3f3f3f;
const int M=200005;
const int N=603;
struct Edge {
	int v;
	Edge *nxt;
}edge[2*M];
Edge *NIL=&edge[0];
Edge *ecnt=&edge[1];
Edge *head[N],*best[N];
Edge *ban;
int n,m;
double ans;
double p[N];
inline void AddEdge(int u,int v) {
	ecnt->v=v;
	ecnt->nxt=head[u];
	head[u]=ecnt++;
}
double Prework(int u) {
	if(p[u]!=-1) return p[u];
	p[u]=0;
	int cnt=0;
	for(Edge *e=head[u];e!=NIL;e=e->nxt)
		cnt++;
	for(Edge *e=head[u];e!=NIL;e=e->nxt) {
		p[u]+=Prework(e->v)+1;
		if(p[e->v]>p[best[u]->v])
			best[u]=e;
	}
	p[u]/=cnt;
	return p[u];
}
double Dfs(int u) {
	if(p[u]!=-1) return p[u];
	p[u]=0;
	int cnt=0;
	for(Edge *e=head[u];e!=NIL;e=e->nxt)
		if(e!=ban)
			cnt++;
	if(cnt==0) return INF;
	for(Edge *e=head[u];e!=NIL;e=e->nxt)
		if(e!=ban)
			p[u]+=Dfs(e->v)+1;
	p[u]/=cnt;
	return p[u];
}
int main() {
	NIL->v=0; NIL->nxt=NIL;
	scanf("%d %d",&n,&m);
	for(int i=0;i<=n;i++) {
		best[i]=head[i]=NIL;
		p[i]=-1;
	}
	for(int i=1,s,t;i<=m;i++) {
		scanf("%d %d",&s,&t);
		AddEdge(s,t);
	}
	p[n]=0;
	ans=Prework(1);
	for(int j=1;j<n;j++) {
		p[n]=0;
		for(int i=1;i<n;i++)
			p[i]=-1;
		ban=best[j];
		ans=min(ans,Dfs(1));
	}
	printf("%.10lf\n",ans);
	return 0;
}