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 <cstdio>
#include <algorithm>
using namespace std;
typedef long double LD;
#define N 600 + 5

int n, m, Go[N][N];
LD ans, PreP[N], Pre[N], Suf[N];

void DP()
{
	PreP[1] = 1, Pre[1] = 0;
	for (int i = 1; i < n; i ++)
	{
		LD tp = PreP[i] / Go[i][0], tpl = (Pre[i] + PreP[i]) / Go[i][0];
		for (int j = 1; j <= Go[i][0]; j ++)
		{
			int d = Go[i][j];
			PreP[d] += tp;
			Pre[d] += tpl;
		}
	}
	Suf[n] = 0;
	for (int i = n - 1; i; i --)
	{
		for (int j = 1; j <= Go[i][0]; j ++)
			Suf[i] += Suf[Go[i][j]] + 1;
		Suf[i] /= Go[i][0];
	}
	ans = Suf[1];
}

int main()
{
	scanf("%d%d", &n, &m);
	for (int i = 1, u, v; i <= m; i ++)
	{
		scanf("%d%d", &u, &v);
		Go[u][++ Go[u][0]] = v;
	}
	DP();
	for (int i = 1; i <= n; i ++)
		if (Go[i][0] > 1)
		{
			LD retpl = Pre[n] - PreP[i] * Suf[i];
			for (int j = 1; j <= Go[i][0]; j ++)
			{
				LD npl = retpl + PreP[i] * (Suf[i] * Go[i][0] - (Suf[Go[i][j]] + 1)) / (Go[i][0] - 1);
				ans = min(ans, npl);
			}
		}
	printf("%.9Lf\n", ans);
	return 0;
}