Problem: 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}.

Examples

Input

4 6
1 4
2 3
1 3
1 2
3 4
2 4


Output

1.5000000000


Input

3 2
1 2
2 3


Output

2.0000000000


Input

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


Output

3.0133333333

Solution (in CPP):

Solution:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<stack>
#include<vector>
#include<map>
using namespace std;
const int maxn = 6e2+10;
const int maxm = 36e4+10;

int read(){
	int x=0,f=1;
	char c=getchar();
	while(!isdigit(c)){if(c=='-')f=-1;c=getchar();}
	while(isdigit(c)){x=(x<<3)+(x<<1)+c-48;c=getchar();}
	return x*f;
}

int to[maxm],net[maxm],h[maxn],e=1;
inline void add(int x,int y){
	to[++e]=y;net[e]=h[x];h[x]=e;
}

int k,m,n;
int siz[maxn];
double ans=1e15;
double dp[maxn];
inline void work(int now){
	dp[n]=0;
	double sum,max_;
	for(register int x=n-1;x>=1;x--){
		sum=0;max_=0;
		for(register int i=h[x];i;i=net[i]){
			int y=to[i];
			sum+=dp[y];
			max_=max(max_,dp[y]);
		}
		if(now==x && siz[x]>=2){
			sum-=max_;
			dp[x]=sum/(siz[x]-1)+1;
		}
		else dp[x]=sum/siz[x]+1;
	}
	ans=min(ans,dp[1]);
}

int main(){
	n=read();m=read();
	for(register int i=1;i<=m;i++){
		int x=read(),y=read();
		siz[x]++;
		add(x,y);
	}
	for(register int i=0;i<n;i++)work(i);
	printf("%.8lf\n",ans);
	return 0;
}