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>
std::vector<int>G[605];
double f[605],ans;
int n,m,vis[605],sel,u,v;
double dfs(int u){
	if (u==n)return 0;
	if (f[u]!=-1)return f[u];
	f[u]=0;
	int max=0,cnt=0;
	for(int i=0;i<G[u].size();++i){
		if (!vis[G[u][i]])continue;
		f[u]+=dfs(G[u][i]);if (f[G[u][i]]>f[max])max=G[u][i];cnt++;
	}
	if (sel!=u)f[u]=f[u]/cnt+1;
	else f[u]=(f[u]-f[max])/(cnt-1)+1;
	return f[u];
}

int work(int u){
	if (vis[u]!=-1)return vis[u];
	vis[u]=0;
	for (int i=0;i<G[u].size();++i)
		if (work(G[u][i])){vis[u]=1;break;}
	return vis[u];
}
	

int main(){
	scanf("%d%d",&n,&m);
	std::memset(vis,-1,sizeof(vis));vis[n]=1;
	for(int i=1;i<=m;++i){
		scanf("%d%d",&u,&v);
		G[u].push_back(v);
	}for(int i=1;i<=n;++i)work(i);
	sel=0;for(int i=1;i<=n;++i)f[i]=-1;ans=dfs(1);
//	for(int i=1;i<=n;++i)printf("%.2f ",f[i]); 
	for(sel=1;sel<=n;++sel){
		if (!G[sel].size())continue;
		for(int i=1;i<=n;++i)f[i]=-1;
		ans=std::min(ans,dfs(1));
	}printf("%.8f",ans);
	return 0;
}