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;

#define int long long
#define rep(i,l,r) for(int i=(int)(l);i<(int)(r);i++)
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)x.size())
template<class T>bool chmax(T &a,T b){if(a<b){a=b;return 1;}return 0;}
template<class T>bool chmin(T &a,T b){if(a>b){a=b;return 1;}return 0;}

typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;

const int inf = 1LL<<60;
const int mod = 1e9 + 7;
const double eps = 1e-9;

/*{
  }*/

int n, m;
vi es[601];

signed main(){
  cin >> n >> m;
  rep(i, 0, m){
    int s, t;
    cin >> s >> t;
    es[s].emplace_back(t);
  }
  
  double ans = inf;
  rep(k, 0, n){
    vector<double> dp(n+1);
    dp[n] = 0;
    for(int i= n-1; i > 0; i--){
      double sum = 0.0;
      double ma = 0.0;
      int num = sz(es[i]);

      for(auto to : es[i]){
        sum += dp[to];
        chmax(ma, dp[to]);
      }

      if(i == k and num >= 2){
        num--;
        sum -= ma;
      }

      dp[i] = sum/num + 1;
    }

    chmin(ans, dp[1]);
  }

  printf("%.9f\n", ans);

  return 0;
}