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;
const int MAXN = 600;
bool A[MAXN + 5][MAXN + 5];
double f[MAXN + 5], g[MAXN + 5], d[MAXN + 5];
int main() {
	int n, m; scanf("%d%d", &n, &m);
	for(int i=1;i<=m;i++) {
		int a, b; scanf("%d%d", &a, &b);
		A[a][b] = true, d[a]++;
	}
	for(int i=n;i>=1;i--) {
		f[i] = 0;
		for(int j=i+1;j<=n;j++)
			if( A[i][j] )
				f[i] += (f[j] + 1) / d[i];
	}
	double ans = f[1];
	for(int i=n;i>=1;i--) {
		if( d[i] > 1 ) {
			int mx = -1;
			for(int j=i+1;j<=n;j++)
				if( A[i][j] ) {
					if( mx == -1 || f[j] > f[mx] )
						mx = j;
				}
			g[i] = 0;
			for(int j=i+1;j<=n;j++)
				if( A[i][j] ) {
					if( j != mx ) {
						g[i] += (g[j] + 1) / (d[i] - 1);
					}
				}
			for(int j=i-1;j>=1;j--) {
				g[j] = 0;
				for(int k=j+1;k<=n;k++)
					if( A[j][k] )
						g[j] += (g[k] + 1) / d[j];
			}
			ans = min(ans, g[1]);
		}
		g[i] = f[i];
	}
	printf("%.7lf\n", ans);
}