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;
using ll = long long;

ll N, M;
ll dels=-1, delt=-1;

double updatedp(vector<vector<ll>> &g,vector<double> &dp,ll v){
	if(v==N-1)return 0.;
	if(dp[v]>-0.1)return dp[v];
	double ans=0.;
	ll gvsize=g[v].size();
	for(ll nextv:g[v]){	
		if(v==dels && nextv==delt){
			gvsize--;
			continue;
		}
		ans+=updatedp(g,dp,nextv);
	}
	return dp[v]=1.+ans/gvsize;
}

signed main() {
  	scanf("%lld%lld", &N, &M);
	vector<vector<ll>> g(N);
	for(ll i=0;i<M;i++){
		ll s,t;
		scanf("%lld%lld", &s, &t);
		s--;t--;
		g[s].push_back(t);
	}

	vector<double> dptmp(N,-1.);
	double ans=updatedp(g,dptmp,0);

	vector<pair<ll,ll>> delpath;
	for(ll v=0;v<N;v++){
		if(g[v].size()==1)continue;
		double maxval=0.;
		ll tmps=-1,tmpt=-1;
		for(ll nextv:g[v]){
			if(maxval<dptmp[nextv]){
				maxval=dptmp[nextv];
				tmps=v;
				tmpt=nextv;
			}
		}
		delpath.push_back({tmps,tmpt});
	}

	for(pair<ll,ll> path:delpath){
		dels=path.first,delt=path.second;
		vector<double> dp(N,-1.);
		double tmp=updatedp(g,dp,0);
		ans=min(ans,tmp);
	}

	printf("%.20f\n", ans);
	return 0;
}