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 pi 3.1415926535897932384626
using namespace std;
      
struct P{
  int x,y;
  bool operator < (const P &a )const{
    //if(y!=a.y)
    return y>a.y;
  };
};

int a,b,c,d,i,k,n,m,e,dx[10]={1,0,-1,0,1,1,-1,-1},dy[10]={0,1,0,-1,1,-1,1,-1};
long double o[613],x;
int l[610];
int j[302];

long long y,z,mod=1000000007;
//string r;
char r[111];

P u[1];
queue<int> q;

stack<int> s;
map<int,int> p;
vector<int> v[666];

bool as(P a,P b)
{
  if(a.x==b.x)
  return a.y<b.y;
  return a.x<b.x;
}

void f(int n)
{
  memset(o,0,sizeof(o));
  for(int t=a-1;t;t--)
  {
    long double y=0;
    for(int h=0;h<v[t].size();h++)
      o[t]+=1+o[v[t][h]],y=max(y,1+o[v[t][h]]);
    if(t==n) o[t]=(o[t]-y)/(v[t].size()-1);
    else o[t]=o[t]/v[t].size();
  }
  //for(int t=1;t<=a;t++)
  //  printf("%Lf ",o[t]);
  x=min(x,o[1]);
}

int main()
{
  
  scanf("%d %d",&a,&b);
  for(int t=1;t<=b;t++)
  {
    scanf("%d %d",&n,&m);
    v[n].push_back(m);
    l[m]=n;
  }
  x=12345678;
  f(0);
  for(int t=1;t<a;t++)
  if(v[t].size()>1) f(t);
  
  printf("%Lf",x);
  
}