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>
using namespace std;
using ld=long double;
typedef pair<int,int> P;
ld dp[605],EPS=1e-15,ans=1e9;
template<typename T>
void chmin(T &a,T b){
  if(a>b)a=b;
}

signed main(){
  int n,m;cin>>n>>m;
  vector<int> edge[n];
  while(m--){
    int s,t;cin>>s>>t;
    edge[--s].push_back(--t);
  }
  ld ans=1e9;
  for(int i=0;i<=n;i++){
    if(edge[i].size()==1)continue;
    memset(dp,0,sizeof(dp));
    for(int j=n-1;j>=0;j--){
      if(i==j){
        ld tmp=0;int a;
        for(int p:edge[j])if(tmp<dp[p])tmp=dp[p],a=p;
        for(int p:edge[j])if(p!=a)dp[j]+=(dp[p]+1)/((ld)edge[j].size()-1);
      }
      else for(int p:edge[j])dp[j]+=(dp[p]+1)/(ld)edge[j].size();
    }
    if(dp[0]>EPS)chmin(ans,dp[0]);
  }
  cout<<fixed<<setprecision(12)<<ans<<endl;
}

/*
解説AC
扉閉める条件なくても期待値求められなくね？
->O(NM)で求める方法は思いついたけどこれ以上どうしろと？
->解説見る
期待値は後ろから求めて行けばよさげ？あと部屋からNにいける道しかないのが楽でよかった
別に後ろからじゃなくてもいいのかな
いやでも前からだと通路通った分の+1が出来ない気がするんだよな
後ろからだと線形になっているからなんとかなるんだけど
後ろから求める方法さえ分かれば最大のそれを削るのは割と明らかに見える
*/