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>

#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;

int main(){
	int n,m; scanf("%d%d",&n,&m);
	vector<vector<int>> G(n);
	rep(i,m){
		int u,v; scanf("%d%d",&u,&v); u--; v--;
		G[u].emplace_back(v);
	}

	double dp0[600]={};
	for(int u=n-2;u>=0;u--){
		for(int v:G[u]){
			dp0[u]+=dp0[v];
		}
		dp0[u]=dp0[u]/G[u].size()+1;
	}

	double ans=1e77;
	rep(i,n){
		int j=-1;
		for(int k:G[i]){ if(j==-1 || dp0[j]<dp0[k]) j=k; }

		double dp[600]={};
		for(int u=n-2;u>=0;u--){
			int deg=0;
			for(int v:G[u]) if(!(u==i && v==j)) {
				dp[u]+=dp[v];
				deg++;
			}
			dp[u]=dp[u]/deg+1;
		}
		ans=min(ans,dp[0]);
	}
	printf("%.15f\n",ans);

	return 0;
}