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 <vector>
#include <algorithm>
#include <math.h>
#include <queue>
using namespace std;
typedef long long int ll;

vector<int> g[650];

int main(){
	cin.tie(nullptr);
	ios::sync_with_stdio(false);
	int n,m; cin >> n >> m;
	for(int i=0;i<m;i++){
		int x,y; cin >> x >> y;
		x--; y--;
		g[x].push_back(y);
	}
	vector<double> dp(n);
	dp[n-1]=0;
	for(int i=n-2;i>=0;i--){
		double now=0;
		for(int p:g[i]){
			now+=dp[p];
		}
		now/=g[i].size();
		dp[i]=now+1;
	}
	double ans=dp[0];
	for(int i=0;i<n-1;i++){
		if(g[i].size()==1)continue;
		int nu=g[i][0];
		for(int p:g[i]){
			if(dp[nu]<dp[p])nu=p;
		}

		double now=[&]{
			vector<double> dp(n);
			dp[n-1]=0;
			for(int j=n-2;j>=0;j--){
				double now=0;
				for(int p:g[j]){
					if(i==j&&p==nu)continue;
					now+=dp[p];
				}
				int deg=g[j].size();
				if(i==j)deg--;
				now/=deg;
				dp[j]=now+1;
			}
			return dp[0];
		}();
		ans=min(ans,now);
	}
	printf("%.8f\n", ans);
}