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: Wrong Answer)

#include <bits/stdc++.h>
 
using namespace std;
typedef long long LL;
typedef pair<int,int> pii;
typedef int ftype;
 
#define pb push_back 
#define eb emplace_back
#define mp make_pair
#define fi first
#define se second 
#define debug(x)  do{cerr<<#x<<": "<<x<<"\n";}while(0)
#define debugv(x) do{cerr<<#x<<": ";for(auto&e: (x))cerr<<e<<" ";cerr<<"\n";}while(0)
#define fori(i, n) for (int i = 0; i < (int)(n); ++i)
#define forn(i, n) for (int i = 1; i <= (int)(n); ++i)
#define ford(i, n) for (int i = (int)(n); i >= 0; --i)
#define fore(i, a, b) for (int i = (int)(a); i <= (int)(b); ++i)
#define RETURN(x) do{cout << x << '\n'; return 0;}while(0)
#define FIRSTWIN do{cout << "Monocarp\n"; return 0;}while(0)
#define SECONDWIN do{cout << "Bicarp\n"; return 0;}while(0)

const int MAXN = 606;
const int MOD = 1e9+7;
const int INF = 1e9+1;
double PI = 3.14159265358979323846;

int N,M;
pii edge[MAXN];
int adj[MAXN][MAXN];
int removed = -1;
int ok[MAXN];
double dp[MAXN];
int B[MAXN];

int isconnect(int u, int v) {
    return adj[u][v] && (removed == -1 || edge[removed] != mp(u,v)); 
}

int can() {
    ok[N] = 1;
    fore(u,1,N) ok[u] = 0;

    for(int u = N-1; u >= 1; u--) {
        fore(v,u+1,N) {
            if(isconnect(u,v)) {
                ok[u] |= ok[v];
            }
        }
    }

    return ok[1];
}

double f(int u) {
    if(u == N) return 0;

    if(B[u]) return dp[u];
    B[u] = 1;
    dp[u] = 0;

    int k = 0;
    fore(v,u+1,N) {
        if(isconnect(u,v)) {
            dp[u] += 1+f(v);
            k++;
        } 
    }

    assert(k);
    dp[u] /= k;
    return dp[u];
}

int main() {
#ifdef OJ
    freopen("input.txt", "rt", stdin);
    //freopen("output.txt", "wt", stdout);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);

    cin >> N >> M;
    fori(i,M) {
        int s,t;
        cin >> s >> t;
        edge[i] = {s,t};
        adj[s][t] = 1;
    }

    if(!can()) {
        cout << 0 << endl;
        return 0;
    }

    double mn = f(1);

    for(removed = 0; removed < M; removed++) {
        if(can()) {
            fore(u,1,N) {
                B[u] = 0;
                dp[u] = 0;
            }
            mn = min(mn, f(1));
        }
    }

    cout << std::fixed << setprecision(8);
    cout << mn << endl;
    return 0;
}