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<bits/stdc++.h>
using namespace std;
#define next Next
#define last Last
const int N=360005;
int n,m;
vector<int>g[N];
double ans,f[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++;}
#define gc getchar
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;
}
void solve(int id)
{
	f[n]=0;
	for(int i=n-1;i;i--)
	{
		int du=g[i].size();
		double sum=0,ma=0;
		for(int j=0;j<g[i].size();j++)
		{
			int v=g[i][j];
			sum+=f[v];
			ma=max(ma,f[v]);
		}
		if(i==id&&du>1)
		{
			sum-=ma;
			du--;
		}
		f[i]=sum/du+1;
	}
	ans=min(ans,f[1]);
}
signed main()
{
	n=read();m=read();
	for(int i=1;i<=m;i++)
	{
		int x=read(),y=read();
		g[x].push_back(y);
	}
	ans=1e9;
	for(int i=0;i<n;i++)solve(i);
	printf("%.8lf\n",ans);
}