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 <vector>
#include <cmath>
#include <algorithm>
#define N 605
#define INF 0x3f3f3f3f
using namespace std;
int n,m;
vector<int>v[N];
double dp[N]={0};
double process(int p)//对于p这个顶点删除其连着的最大的出边 
{
	int i,j;
	for(i=1;i<=n;i++)dp[i]=0.0;
	dp[n]=0.0;
	for(i=n-1;i>=1;i--)
	{
		if(v[i].size()==1&&p==i)//i出度为1且 要删它的边 
		{
			dp[i]=INF;
			continue;
		}
		double len=v[i].size()-(p==i);
		dp[i]=1.0;
		double mmax=0;
		for(j=0;j<v[i].size();j++)
		{
			double temp=dp[v[i][j]];
			mmax=max(mmax,temp);
			dp[i]+=temp*1.0/len;
		}
		if(i==p)dp[i]-=mmax*1.0/len;
	}
	//cout<<dp[1]<<endl;
	return dp[1]; 
}
int main()
{
	cin>>n>>m;
	int i;
	for(i=1;i<=m;i++)
	{
		int s,t;
		scanf("%d%d",&s,&t);
		 v[s].push_back(t);//出边 
	}
	double ans=INF;
	for(i=n;i>=1;i--)
	{
		ans=min(ans,process(i));
	}
	printf("%.10lf",ans);
	return 0;
}