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<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<vector>
#include<queue>
#include<cmath>
#define ll long long
#define F(i,a,b) for(int i=(a);i<=(b);i++)
using namespace std;
template<class T>inline void read(T &x) {
    x=0; int ch=getchar(),f=0;
    while(ch<'0'||ch>'9'){if (ch=='-') f=1;ch=getchar();}
    while (ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
    if(f)x=-x;
}
const int inf=0x3f3f3f3f;
const int maxn=650;
int n,m;
double f[maxn];
vector<int> e[maxn];
double ans=inf;
void calc(int p)
{
	for(int i=0;i<=n;i++) f[i]=0;
	for(int i=n-1;i>=1;i--)
	{
		int num=e[i].size();
		double maxx=0.0,sum=0.0;
		for(int j=0;j<e[i].size();j++)
		{
			int v=e[i][j];
			sum+=f[v];
			maxx=max(maxx,f[v]);
		}
		
		if(i==p&&num>1) sum-=maxx,num--;
		
		f[i]=(sum/(double)num)+1;
	}
	ans=min(ans,f[1]);
}
int main()
{
	read(n),read(m);
	for(int i=1;i<=m;i++)
	{
		int x,y;read(x),read(y);
		e[x].push_back(y);
	}
	for(int i=1;i<=n-1;i++) calc(i);
	printf("%.8lf\n",ans);
	return 0;
}