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;

template<typename T>
void out(T x) { cout << x << endl; exit(0); }
#define watch(x) cout << (#x) << " is " << (x) << endl





using ll = long long;


const int maxn = 605;

int n, m;
vector<int> g[maxn];


double solve(int block) {
    vector<double> dp(n+10);
    dp[n] = 0;
    
    for (int at=n-1; at>=1; at--) {
	int outd = 0;
	double sum = 0;
	double hi = 0;
	for (int to: g[at]) {
	    outd++;
	    sum += dp[to];
	    hi = max(hi, dp[to]);
	}

	if (at == block && outd>=2) {
	    sum -= hi;
	    outd--;
	}

	dp[at] = 1 + sum/outd;
    }

    return dp[1];
}

int main() {
    ios_base::sync_with_stdio(false); cin.tie(0);  cout.tie(0);

    cin>>n>>m;
    for (int i=0; i<m; i++) {
	int u,v; cin>>u>>v;
	g[u].push_back(v);
	assert(u<v);
    }



    double res = 1e18;
    for (int i=0; i<=n-1; i++) {
	res = min(res, solve(i));
    }

    cout<<fixed<<setprecision(15)<<res<<endl;    
    return 0;
}