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;
typedef long long ll;
typedef pair<int,int> ii;

vector<int> adj[605];
vector<ii> edges;
double sum[605], p[605], e[605];


int main(){

    int n, m; cin >> n >> m;
    for(int i = 0; i < m; i++){
        int u, v; cin >> u >> v;
        edges.push_back({u, v});
        adj[u].push_back(v);
    }
    
    p[1] = 1;
    for(int u = 1; u <= n; u++)
        for(int v : adj[u])
            p[v] += p[u]/adj[u].size();
    e[n] = 1;
    for(int u = n; u >= 1; u--){
        for(int v : adj[u])
            sum[u] += e[v];
        if(adj[u].size())
            e[u] = sum[u]/adj[u].size() + 1;
    }
    
    double res = e[1];
    // printf("%.6f\n", res);
    // return 0;

    for(ii par : edges){
        int u = par.first, v = par.second;
        if(adj[u].size() == 1) continue;
        double aux = (sum[u] - e[v])/(adj[u].size() - 1) + 1;
        double delta = e[1] + (aux - e[u])*p[u];
        res = min(res, delta);
    }
    printf("%.6f\n", res-1);
	return 0;
}