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 <stdio.h>
#include <vector>
using namespace std;
int n, m, s, t;
vector<int>g[600];
int d[600];
double e[600];
double l;
int main() {
	scanf("%d%d", &n, &m);
	for (int i = 0; i < m; i++) {
		scanf("%d%d", &s, &t);
		g[s - 1].push_back(t - 1);
	}
	for (int i = 0; i < n; i++) {
		d[i] = g[i].size();
	}
	for (int i = n - 2; i >= 0; i--) {
		l = 0;
		for (int j = 0; j < d[i]; j++) {
			if (e[g[i][j]] > l)l = e[g[i][j]];
			e[i] += e[g[i][j]] / d[i];
		}
		for (int j = 0; j < d[i]; j++) {
			if (e[g[i][j]] == l) {
				t = g[i][j];
				g[i][j] = g[i][d[i] - 1];
				g[i][d[i] - 1] = t;
				break;
			}
		}
		e[i]++;
	}
	l = e[0];
	for (int i = 0; i < n - 1; i++) {
		if (d[i] == 1)continue;
		e[n - 1] = 0;
		d[i]--;
		for (int j = n - 2; j >= 0; j--) {
			e[j] = 0;
			for (int k = 0; k < d[j]; k++) {
				e[j] += e[g[j][k]] / d[j];
			}
			e[j]++;
		}
		if (l > e[0])l = e[0];
		d[i]++;
	}
	printf("%.15lf\n", l);
}