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;
#define rep(i,n) for(int i=0; i<(n); i++)
const int MX = 605;
int n, m, s, t, sz[MX];
vector<int> G[MX], rG[MX];
double p[MX], d[MX], ans;
int main(){
	scanf("%d%d", &n, &m);
	rep(i,m){
		scanf("%d%d", &s, &t);
		s--; t--;
		sz[s]++;
		G[s].push_back(t);
		rG[t].push_back(s);
	}
	for(int i = n-1; i >= 0; i--){
		if(i != n-1) p[i] += 1;
		for(int pr: rG[i]) p[pr] += p[i] / sz[pr];
	}
	for(int i = n-2; i >= 0; i--){
		double dd[MX] = {};
		dd[i] = 1;
		for(int j = i; j >= 0; j--){
			for(int pr: rG[j]) dd[pr] += dd[j] / sz[pr];
		}
		d[i] = dd[0];
	}
	ans = p[0];
	rep(i,n) if(sz[i] > 1){
		for(int nx: G[i]){
			double np = ((p[i] - 1) * sz[i] - p[nx]) / (sz[i]-1) + 1;
			ans = min(ans, p[0] + (np - p[i]) * d[i]);
		}
	}
	printf("%.9f\n", ans);
}