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<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN=1e3+5;
const int MAXM=6e5+6;
int edge,head[MAXN],tail[MAXM],nex[MAXM];
int n,m;
double ans=1e15,dp[MAXN];
void add(int u,int v){
	edge++,nex[edge]=head[u],head[u]=edge,tail[edge]=v;
}
int main(){
	scanf("%d%d",&n,&m);
	for (int i=1;i<=m;i++){
		int u,v; scanf("%d%d",&u,&v);
		add(u,v);
	}
	for (int i=0;i<n;i++){
		dp[n]=0;
		for (int j=n-1;j>=1;j--){
			double num=0,sum=0,mx=0;
			for (int e=head[j];e;e=nex[e]){
				int v=tail[e];
				sum+=dp[v];
				mx=max(mx,dp[v]);
				num++;
			}
			if (j==i&&num>1){
				num--; sum-=mx;
			}
			dp[j]=sum/num+1;
		}
		ans=min(ans,dp[1]);
	}
	printf("%.6lf\n",ans);
	return 0;
}