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;
typedef long long ll;
#define pb push_back
#define fi first
#define se second
const int maxn=605;
const int mod=1e9+7;
const int inf=0x3f3f3f3f;
vector<int> g[maxn];
long double dp[maxn];
int n,m;
long double cal(int x)
{
    for(int i=0;i<=n;i++) dp[i]=0;
    for(int i=n-1;i>=1;i--)
    {
        long double sum=0,Max=0;
        for(int j=0;j<g[i].size();j++) sum+=dp[g[i][j]],Max=max(Max,dp[g[i][j]]);
        if(i==x&&g[i].size()>=2)
        {
            sum-=Max;
            dp[i]=sum/(g[i].size()-1);
        }
        else dp[i]=sum/g[i].size();
        dp[i]+=1;
    }
    return dp[1];
}
int main()
{
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    cin>>n>>m;
    for(int i=0;i<m;i++)
    {
        int u,v;
        cin>>u>>v;
        g[u].pb(v);
    }
    long double ans=cal(1);
    //cout<<ans<<'\n';
    for(int i=2;i<=n-1;i++)ans=min(ans,cal(i));
    cout<<fixed<<setprecision(8)<<ans<<'\n';
    return 0;
}