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 <utility>
using namespace std;
vector<int> G[610];
long double dp[610],ans = 6000000,INF = 1000000000000;
int a[610];
pair<int,int> p[200010];
int main(){
	cout.precision(20);
	int i,j,n,m;
	cin >> n >> m;
	for(i=0;i<m;i++){
		int s,t;
		cin >> s >> t; s--; t--;
		G[s].push_back(t);
		p[i] = {s,t};
	}
	for(i=0;i<n;i++){
		dp[i] = INF;
	}
	dp[n - 1] = 0;
	for(i=n - 2;i>=0;i--){
		long double mx = -1,sum = 0;
		j = -1;
		for(int v:G[i]){
			sum += dp[v];
			if(dp[v]>mx){
				mx = dp[v];
				j = v;
			}
			a[i] = j;
		}
		long double sz = G[i].size();
		dp[i] = sum/sz + (long double)1;
	}
	ans = min(ans,dp[0]);
	for(i=0;i<n;i++){
		for(j=0;j<n;j++){
			dp[j] = INF;
		}
		dp[n - 1] = 0;
		for(j=n - 2;j>=0;j--){
			long double sum = 0;
			for(int v:G[j]){
				if(j==i && v==a[j]) continue;
				sum += dp[v];
			}
			long double sz = G[j].size();
			if(j==i) sz--;
			dp[j] = sum/sz + (long double)1;
		}
		ans = min(ans,dp[0]);
	}
	cout << ans << endl;
}