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<algorithm>
#include<cmath>
#include<vector>
using namespace std;
#define LL long long
#define MAXN 600
#define INF 100000000000000000
#define DB double
#define PI acos(-1.0)
int n,m;
int u[MAXN+5],v[MAXN+5],tmp;
DB f[MAXN+5],ans;
vector<int> G[MAXN+5];
LL read(){
  LL x=0,F=1;char c=getchar();
  while(c<'0'||c>'9'){if(c=='-')F=-1;c=getchar();}
  while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
  return x*F;
}
DB del(int x){
  f[n]=1;
  for(int i=n-1;i>=1;i--){
      f[i]=0;
      int siz=G[i].size(),id=i;
      DB mx=0;
      for(int k=0;k<siz;k++){
          int pre=G[i][k];
          f[i]+=f[pre];
          mx=max(mx,f[pre]);
      }
      if(siz>=2&&x==i){
        f[i]-=mx;
        siz--;
      }
      f[i]=f[i]/siz+1;
  }
  return f[1]-1;
}
int main()
{
  n=read(),m=read();tmp=n+1;
  for(int i=1;i<=m;i++){
    int a=read(),b=read();
    G[a].push_back(b);
  }
  ans=del(0);
  for(int i=1;i<n;i++)
  ans=min(ans,del(i));
  printf("%.10f",ans);
}