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 <algorithm>
#include <stdio.h>
using namespace std;
int n,m;
int s[200000],t[200000];
vector<int> v[700];
double e[700],ans,ma;
int main(){
    cin>>n>>m;
    for(int i=0;i<m;i++){
        cin>>s[i]>>t[i];
        v[s[i]].push_back(t[i]);
    }
    for(int i=n-1;i;i--){
        for(int j=0;j<v[i].size();j++){
            e[i]+=(1.0+e[v[i][j]])/v[i].size();
        }
    }
    ans=e[1];
    for(int i=1;i<n;i++){
        if(v[i].size()==1)continue;
        ma=-1.0;
        e[i]=0.0;
        for(int j=0;j<v[i].size();j++){
            if(ma<e[v[i][j]]){
                e[i]+=(ma+1.0)/(v[i].size()-1.0);
                ma=e[v[i][j]];
            }else e[i]+=(e[v[i][j]]+1.0)/(v[i].size()-1.0);
        }
        for(int k=i-1;k;k--){
            e[k]=0.0;
            for(int j=0;j<v[k].size();j++){
                e[k]+=(1.0+e[v[k][j]])/v[k].size();
            }
        }
        ans=min(ans,e[1]);
    }
    printf("%.10f\n",ans);
    return 0;
}