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;

const int N = 606;
vector < int > g[N];
int memo[N][N], n;
double dp[N][N];

double solve(int blocked, int u){
  double resp = .0;
  if(u == n) return .0;
  double mx = -1;
  int mx_vtx;
  if(g[u].size() == 0) return 1e9;
  if(memo[blocked][u] != -1) return dp[blocked][u];
  memo[blocked][u] = 1;
  for (int v : g[u]){
    double aux = solve(blocked, v);
    resp += aux;
    if(mx < aux) {
      mx = aux, mx_vtx = v;
    }
  }
  if(u != blocked){
    return dp[blocked][u] = (resp/(double)g[u].size()) + 1;
  }
  else {
    resp -= solve(blocked, mx_vtx);
    if(g[u].size() == 1) return dp[blocked][u] = 1e9;
    return dp[blocked][u] = (resp/((double)g[u].size()-1)) + 1;
  }
}

int main(){
  int m;
  scanf("%d %d", &n, &m);
  for (int i=0; i<m; i++){
    int u, v;
    scanf("%d %d", &u, &v);
    g[u].push_back(v);
  }
  memset(memo, -1, sizeof memo);
  double ans = solve(0, 1);
  for (int i=1; i<=n; i++){
    ans = fmin(ans, solve(i, 1));
  }
  printf("%.8lf\n", ans);
}