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>
#define mz 1000000007

using namespace std;

double dp[606][605];
int cnt[606];
bool life[606][606];

int main()
{
    int n,m,x,y,now=1;
    double maxx,minn;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++)
    {
        scanf("%d%d",&x,&y);
        cnt[x]++;
        life[x][y]=1;
    }
    for(int i=n-1;i>=1;i--)
    {
        maxx=-1;
        minn=1;
        for(int j=i+1;j<=n;j++)
            if(life[i][j])
            {
                dp[i][0]+=dp[j][0];
                maxx=max(maxx,dp[j][0]);
                for(int k=i+1;k<=n;k++)
                    dp[i][k]+=dp[j][k];
            }
        for(int j=1;j<=i;j++)
            dp[i][j]=dp[i][0]/cnt[i]+1;
        for(int j=i+1;j<=n;j++)
            dp[i][j]=dp[i][j]/cnt[i]+1;
        if(cnt[i]!=1)
            dp[i][i]=(dp[i][0]-maxx)/(cnt[i]-1)+1;

        dp[i][0]=dp[i][0]/cnt[i]+1;
        //cout<<i<<":"<<dp[i][0]<<" "<<dp[i][1]<<endl;
    }
    for(int i=1;i<=n;i++)
        dp[1][0]=min(dp[1][0],dp[1][i]);
    printf("%.8f\n",dp[1][0]);
	return 0;
}