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: Accepted)

#include <stdio.h>
#include <stdlib.h>
#define ll long long
#define rep(i,l,r)for(ll i=(l);i<(r);i++)
#define max(p,q)((p)>(q)?(p):(q))
#define min(p,q)((p)<(q)?(p):(q))

//辺の情報を個別に持つタイプ
typedef struct edge{ll s,g;}E;
typedef struct graph{
	int vcnt,ecnt;
	E  e[200010];//適宜変える(ecnt)
	int id[100010];//適宜変える(vcnt)
}G;
G g;

int esort(const void*a,const void*b){
	E*p=(E*)a,*q=(E*)b;
	if((*p).s<(*q).s)return -1;
	if((*p).s>(*q).s)return  1;
	if((*p).g<(*q).g)return -1;
	return 1;
}
void readgraph(){
	//適宜変える
	ll n,m;
	scanf("%lld%lld",&n,&m);
	rep(i,0,m){
		ll x,y,c;
		scanf("%lld%lld",&x,&y);
		x--,y--;
		g.e[i].s=x;
		g.e[i].g=y;
	}
	g.vcnt=n;
	g.ecnt=m;
	qsort(g.e,g.ecnt,sizeof(E),esort);

	int p=0;
	rep(i,0,g.vcnt){
		while(p<g.ecnt&&g.e[p].s<i)p++;
		g.id[i]=p;
	}
	g.id[g.vcnt]=g.ecnt;//番兵
}

double dp[610];
double dpmax[610];	
double p[610];

int main(){
	readgraph();
	for(int v=g.vcnt-2;v>=0;v--){
		double s=0,cnt=0,M=0;
		for(int i=g.id[v];i<g.id[v+1];i++){
			s+=dp[g.e[i].g];
			cnt++;
			M=max(M,dp[g.e[i].g]);
		}
		dp[v]=s/cnt+1;
		dpmax[v]=M;
	}
	p[0]=1;
	for(int v=0;v<g.vcnt-1;v++){
		int cnt=g.id[v+1]-g.id[v];
		for(int i=g.id[v];i<g.id[v+1];i++)p[g.e[i].g]+=p[v]/cnt;
	}
		
	double ans=dp[0];
	for(int v=0;v<g.vcnt-1;v++){
		int cnt=g.id[v+1]-g.id[v];
		if(cnt>1){
			double diff=dp[v]-(((dp[v]-1)*cnt-dpmax[v])/(cnt-1)+1);
			ans=min(ans,dp[0]-diff*p[v]);
		}
	}
	
	printf("%.9f",ans);
}