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 FOR(i,a,b) for(int i=a;i<=b;i++)
using namespace std;

const int maxn = 10050;
typedef long long ll;

int n,m;
double dp[maxn];
vector<int>G[maxn];

double slove(int p){
    dp[n]=0;
    for(int i=n-1;i>=1;i--){
        int all=G[i].size();
        if(!all||(i==p&&all==1)){dp[i]=0x3f3f3f3f;continue;};
        double MAX=0,ret=0;
        for(int j=0;j<all;j++){
            MAX=max(dp[G[i][j]]+1,MAX);
            ret+=1/double(all-(i==p))*(dp[G[i][j]]+1);
        }
        if(i==p)ret-=1/double(all-1)*MAX;
        dp[i]=ret;
    }
    return dp[1];
}

int main(){
    cin>>n>>m;
    for(int i=1;i<=m;i++){
        int s,t;
        scanf("%d%d",&s,&t);
        G[s].push_back(t);
    }
    double ans=slove(0);
    for(int i=1;i<=n-1;i++){
        ans=min(ans,slove(i));
    }
    printf("%.10f",ans);
}