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:
#pragma GCC optimize("Ofast,unroll-loops")
#include<bits/stdc++.h>
#define ll long long
#define F first
#define S second
#define P pair
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define rep(i,a,b) for(int i=a;i<b;i++)
#define V vector
#define RE return
#define ALL(a) a.begin(),a.end()
#define MP make_pair
#define PB push_back
#define PF push_front
#define FILL(a,b) memset(a,b,sizeof(a))
using namespace std;
double dp[605],p[605];
bool v[605][605];
int n,l[605]; 
int main(){
	int m;
	scanf("%d%d",&n,&m);
	int x,y;
	FOR(i,1,m){
		scanf("%d%d",&x,&y);
		v[x][y]=1;
		l[x]++;
	}
	for(int i=n-1;i>=1;i--){
		FOR(j,i+1,n){
			if(v[i][j])dp[i]+=(dp[j]+1)/l[i];
		}
	}
	double ans=dp[1];
	for(int k=n-1;k>=1;k--){
		if(l[k]==1){
			p[k]=dp[k];continue;
		}
		int maxi=0;
		FOR(i,k+1,n){
			if(dp[i]>dp[maxi]&&v[k][i])maxi=i;
		}
		p[k]=0;
		FOR(i,k+1,n){
			if(i!=maxi&&v[k][i]){
				p[k]+=(dp[i]+1)/(l[k]-1);
			}
		}
		for(int i=k-1;i>=1;i--){
			p[i]=0;
			FOR(j,i+1,n){
				if(v[i][j])p[i]+=(p[j]+1)/l[i];
			}
		}
		ans=min(ans,p[1]);
		p[k]=dp[k];
	}
	printf("%.10f",ans);
	RE 0;
}