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;
double ans=999999999;
double inf=999999999;
int n,m;
double dp[601];
vector<int>e[601];
void get_dp(int x)
{
	memset(dp,0,sizeof(dp));
	dp[n]=0;
	for(int i=n-1;i>=1;i--)
	{
		int tot=0;
		double mx=0;
		for(int j=0;j<e[i].size();j++)
			tot++,mx=max(mx,dp[e[i][j]]),dp[i]+=dp[e[i][j]];
		if(i==x)
		{
			if(tot==1)
				dp[i]=inf;
			else if(tot>=2)
				dp[i]-=mx,dp[i]/=tot-1,dp[i]++;
			else
				dp[i]=inf;
		}
		else if(tot)
			dp[i]/=tot,dp[i]++;
		else
			dp[i]=inf;
	}
	ans=min(ans,dp[1]);
}
signed main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		if(u>v) swap(u,v);
		e[u].push_back(v);
	}
	for(int i=1;i<=n;i++)
		get_dp(i);
	printf("%lf",ans);
	return 0;
}
/*
5 5
1 2
1 3
2 3
3 5
2 5
*/