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 maxn 600
#define eps 1e-8
struct Edge{
	int from,next,to;
}edge[maxn*maxn/2+5];
int fi[maxn+5],se,si[maxn+5];
inline void add_edge(int u,int v){
	edge[++se].next=fi[u],edge[se].to=v,fi[u]=se,si[u]++,edge[se].from=u;
}
double p1[maxn+5],e1[maxn+5],e2[maxn+5];
int main(){
	int n,m,u,v;scanf("%d%d",&n,&m);
	for(int i=0;i<m;i++)scanf("%d%d",&u,&v),add_edge(u,v);
	p1[1]=1;
	for(int i=1;i<=n;i++){
		if(p1[i]<eps)continue;
		e1[i]/=p1[i];
		for(int j=fi[i];j;j=edge[j].next)p1[edge[j].to]+=p1[i]/si[i],e1[edge[j].to]+=p1[i]*(e1[i]+1)/si[i];
	}
	double ans=e1[n],ans1=0;
	for(int i=n-1;i;i--){
		for(int j=fi[i];j;j=edge[j].next){
			e2[i]+=(e2[edge[j].to]+1)/si[i];
		}
	}
	for(int i=1,u,v;i<=se;i++){
		if(si[u=edge[i].from]==1)continue;
		v=edge[i].to;
		ans1=max(ans1,p1[u]*(e2[u]-(e2[u]-(e2[v]+1)/si[u])*si[u]/(si[u]-1)));
	}
	printf("%.10lf",ans-ans1);
	return 0;
}