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;
const int N=605;
int n,m,x,y,a[N][N],d[N];
double f[N],g[N];
signed main(){
	scanf("%d%d",&n,&m);
	while (m--){
		scanf("%d%d",&x,&y);
		a[x][++a[x][0]]=y;
	}
	f[0]=0;
	for (int i=n-1;i;i--){
		for (int j=1;j<=a[i][0];j++)f[i]+=f[a[i][j]]/a[i][0];
		f[i]++;
	}
	double ans=f[1];
	for (int i=1;i<n;i++)
		if (a[i][0]>1){
			int flag=a[i][1];
			for (int j=2;j<=a[i][0];j++)
				if (f[a[i][j]]>f[flag])flag=a[i][j];
			for (int j=1;j<=n;j++)d[j]=a[j][0];
			d[i]--;
			memset(g,0,sizeof g);
			for (int j=n-1;j;j--){
				for (int k=1;k<=a[j][0];k++)
					if (j!=i||a[j][k]!=flag)g[j]+=g[a[j][k]]/d[j];
				g[j]++;
			}
			ans=min(ans,g[1]);
		}
	printf("%.10lf\n",ans);	
}