Problem:
Score : 600 points

Problem Statement

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}.

Sample Input 1

4 6
1 4
2 3
1 3
1 2
3 4
2 4

Sample Output 1

1.5000000000

If Aoki blocks the passage from Room 1 to Room 2, Takahashi will go along the path 1 → 3 → 4 with probability \frac{1}{2} and 1 → 4 with probability \frac{1}{2}. E = 1.5 here, and this is the minimum possible value of E.

Sample Input 2

3 2
1 2
2 3

Sample Output 2

2.0000000000

Blocking any one passage makes Takahashi unable to reach Room N, so Aoki cannot block a passage.

Sample Input 3

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

Sample Output 3

3.0133333333

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

/**
 * code generated by JHelper
 * More info: https://github.com/AlexeyDmitriev/JHelper
 * @author
 */

#include <iostream>
#include <fstream>

#include "bits/stdc++.h"

#define REP(i,x,y) for(int i=(x);i<(y);i++)
#define RREP(i,x,y) for(int i=(y)-1;i>=(x);i--)
#define all(x) (x).begin(),(x).end()
//#define int long long

using namespace std;

// conversion
inline int toInt(string s) {int v; istringstream sin(s);sin>>v;return v;}
template<class T> inline string toString(T x) {ostringstream sout;sout<<x;return sout.str();}

#define dump(x)  cerr << #x << " = " << (x) << endl
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl

typedef long long ll;

const ll MOD = 1e9 + 7;

class FForkInTheRoad {
public:
	void solve(std::istream& in, std::ostream& out) {
        int N, M;
        in >> N >> M;
        vector<int> s(M), t(M);
        REP(i,0,M) in >> s[i] >> t[i];

        vector<vector<int>> st(N+1), ts(N+1);
        REP(i,0,M) {
            st[s[i]].push_back(t[i]);
            ts[t[i]].push_back(s[i]);
        }
        // calculate mean distances from each locations
        vector<double> dp(N+1, 0);
        RREP(current, 1, N) {
            for (const auto next : st[current]) {
                dp[current] += dp[next] + 1;
            }
            dp[current] /= st[current].size();
        }
        // find block candidates
        vector<pair<int,int>> blockCandidates = {{-1, -1}};
        REP(current,1,N) {
            if (st[current].size() == 1) continue;
            double candidate = 0;
            for (const auto next : st[current]) {
                if (dp[candidate] < dp[next]) candidate = next;
            }
            blockCandidates.push_back({current, candidate});
        }
        // calculate mean distances again for all block candidates
        double answer = 1e5;
        for (const auto& block : blockCandidates) {
            vector<double> dp(N+1, 0);
            RREP(current, 1, N) {
                for (const auto next : st[current]) {
                    if (block.first == current && block.second == next) continue;
                    dp[current] += dp[next] + 1;
                }
                if (block.first == current) {
                    dp[current] /= st[current].size() - 1;
                } else {
                    dp[current] /= st[current].size();
                }
            }
            answer = min(answer, dp[1]);
        }
        out << fixed << setprecision(10) << answer << "\n";
	}
};


int main() {
	FForkInTheRoad solver;
	std::istream& in(std::cin);
	std::ostream& out(std::cout);
	solver.solve(in, out);
	return 0;
}