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 JAVA):

Solution:
import java.util.*;
import static java.lang.Math.*;
import java.math.BigInteger;

public class Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		
		// 入力
		int n = sc.nextInt();
		int m = sc.nextInt();
		ArrayList<Integer>[] list = new ArrayList[n+1];
		for(int i = 1; i <= n; i++){
		    list[i] = new ArrayList<Integer>();
		}
		for(int i = 0; i < m; i++){
		    int from  = sc.nextInt();
		    int to = sc.nextInt();
		    list[from].add(to);
		}
		
		// 計算
		double result = Double.MAX_VALUE;
		for(int k = 0; k <= n-1; k++){
		    double[] dp = new double[n+1];
		    for(int i = n-1; i >= 1; i--){
		        if(i == k && list[i].size() <= 1){
		            dp[i] = 0;
		            continue;
		        }
		        int cnt = list[i].size();
		        if(i == k) cnt--;
		        double q = (double)1/cnt;
		        double max = 0;
		        for(int to : list[i]){
		            if(dp[to] == 0 && to != n){
		                dp[i] = 0;
		                break;
		            }else{
		                dp[i] += (1 + dp[to]) * q;
		                max = max(max, dp[to]);
		            }
		        }
		        if(i == k) dp[i] -= (1 + max) * q;
		        //System.out.println("k:" + k + " " + Arrays.toString(dp));
		    }
		    //System.out.println("k:" + k + " " + Arrays.toString(dp));
		    if(dp[1] != 0) result = min(result, dp[1]);
		}
		// 出力
		System.out.println(result);
		
	}
}