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;
#define pp pair<int,int>
#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define ll long long
#define ld long double
#define all(a) (a).begin(),(a).end()
#define mk make_pair
ll MOD=1000000007;
ll mod=998244353;
int inf=1000001000;
ll INF=100000000000000000;


int main() {
    int n,m;
    cin >> n >> m;
    vector<vector<int>> e(n+1,vector<int>(0));
    ld ans;
    vector<ld> d(n+1,0);
    rep(i,m){
        int f,g;
        cin >> f >> g;
        e[f].push_back(g);
    }
    for (int i=n-1;i>0;i--){
        ld c=e[i].size();
        ld h=0;
        rep(j,e[i].size()){
            h+=1+d[e[i][j]];
        }
        d[i]=h/c;
    }
    ans=d[1];
    for (int k=1;k<n;k++){
        if (e[k].size()<2) continue;
        for (int i=n-1;i>0;i--){
        ld c=e[i].size(),p=0;
        if (k==i) c-=1;
        ld h=0;
        rep(j,e[i].size()){
            h+=1+d[e[i][j]];
            p=max(p,d[e[i][j]]+1);
        }
        if (k==i) h-=p;
        d[i]=h/c;
    }
    ans=min(ans,d[1]);
    }
    cout << fixed << setprecision(10) << ans << endl;
}