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 pii pair<int, int>;
#define ss second
#define ff first
#define all(x) ((x).begin(), (x).end())

using namespace std;
using ll = long long;
using ld = long double;

const int mod = 1e9 + 7;
const ll oo = 1e18;

ld ans = (ld)oo;
int n, m;
vector<int> adj[650];
vector<ld> dp(650, 0);

void funcao(int block){
    dp[n] = 0.0;
    for (int i=n-1; i>=1; i--){
        ld sum = 0.0, maxi = 0.0;
        ld num = adj[i].size();
        for (auto x : adj[i]){
            maxi = max(maxi, dp[x]);
            sum += dp[x];
        }

        if (i == block and num >= 2.0){
            sum -= maxi;
            num--;
        }

        dp[i] = sum/num + 1.0;
    }
    ans = min(dp[1], ans);
}

int main(void){
    cin >> n >> m;

    for (int i=0; i<m; i++){
        int a, b;
        cin >> a >> b;
        adj[a].push_back(b);
    }

    for (int i=0; i<n; i++) funcao(i);

    cout << fixed << setprecision(10) << ans << endl;
return 0;
}