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:
#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <map>

#define M 605

using namespace std;

int n, m;
double f[M], a[M], ans;
vector<int>G[M];

double dfs(int u)
{
	if (u == n)
		return 0;
	if (f[u] != 0)
		return f[u];
	for (int v : G[u])
		f[u] += dfs(v);
	return f[u] = f[u] / G[u].size() + 1;
}

int main()
{
	cin >> n >> m;
	for (int i = 1; i <= m; i++)
	{
		int x, y;
		cin >> x >> y;
		G[x].push_back(y);
	}
	dfs(1);
	a[1] = 1;
	for (int i = 1; i < n; i++)							//经过第i个点的概率
		for (int j : G[i])
			a[j] += a[i] / G[i].size();
	for (int i = 1; i < n; i++)
		if (G[i].size() > 1)
		{
			double tmp = 0;
			for (int j : G[i])
				tmp += f[j];
			for (int j : G[i])
				ans = max(ans, a[i] * (f[i] - ((tmp - f[j]) / (G[i].size() - 1) + 1)));
		}
	printf("%.10f\n", f[1] - ans);

	return 0;
}