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 ll  long long int
#define mp make_pair
#define fi first
#define se second
#define pii pair<ll,ll>
#define pb push_back
#define io ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
ll N,M;
double dp[650];
vector<ll> paths[650];
double ans;
void input()
{
	
	cin>>N>>M;
	for(ll i=1;i<=M;i++)
	{
		ll s,t;
		cin>>s>>t;
		paths[s].pb(t);
	}
}
void f(ll block)
{
	dp[N]=0;
	for(ll i=N-1;i>=1;i--)
	{
		double sum=0;
		double mx=0;
		ll num=paths[i].size();
		for(auto to:paths[i])
		{
			sum+=dp[to];
			if(dp[to]>mx)
			{
				mx=dp[to];
			}
		}
		if(block==i&&num>=2)
		{
			sum-=mx;
			num--;
		}
		dp[i]=sum/num+1;
	}
	if(dp[1]<ans)
	{
		ans=dp[1];
	}
	return ;
}
void solve()
{
	input();
	ans=1e18;
	for(ll i=0;i<=N-1;i++)
	{
		f(i);
	}
	cout<<setprecision(10)<<ans<<"\n";
	return ;
}
int main()
{
	solve();
	return 0;
}