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:
/*input
7 8
1 2
1 3
2 4
3 5
3 6
4 7
5 7
6 7
*/
#include <bits/stdc++.h>
#define up(i,a,b) for(int (i) = (a);(i)<=(b);++(i))
#define down(i,b,a) for(int (i) = (b);i>=(a);--i)
#define bits(x,i) ((x >> i) & 1)
#define mid ((l+r)/2)
#define pr pair<int,int>
using namespace std;
const int N = 605;
long double E[N];
bool ok[N];
vector<int> G[N];
int n,m;
long double ans = 1005;
void solve(int block){
	E[n] = 0;
	for(int u=n-1;u>=1;--u){
		long double sum = 0, maxChild = 0;
		int num = G[u].size();
		for (auto v : G[u]){
			sum += E[v];
			maxChild = max(maxChild, E[v]);
		}
		if (u == block && num >= 2){
			sum-=maxChild;
			num--;
		}
		E[u] = sum/num + 1;
	}
	ans = min(ans, E[1]);
}
int main(){
	ios_base::sync_with_stdio(0); cin.tie(0);
	cin >> n >> m;
	up(i,1,m){
		int u,v; cin >> u >> v;
		G[u].push_back(v);
	}
	for(int i=0;i<n;++i) solve(i);
	cout << fixed << setprecision(6) << ans;
	// dfs(1);
}