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>
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define per(i,a,b) for(int i=(a);i>=(b);--i)
using namespace std;

const int N = 600 + 5;
int mp[N][N], n, m;
pair<int, int> node[N * N];
double ans = INT_MAX;
inline void calc (int id) {
	static double dp[N];
	dp[n] = 0;
	per (i, n - 1, 1) {
		dp[i] = 0;
		double mx = -1;
		rep (j, 1, mp[i][0]) {
			int v = mp[i][j];
			if (dp[v] < 0) { dp[i] = -1; break; }
			dp[i] += dp[v];
			mx = max (mx, dp[v]);
		}
		if (dp[i] < 0) continue;
		if (mp[i][0] - (i == id) == 0) dp[i] = -1;
		else {
			if (i == id) dp[i] -= mx;
			dp[i] = (dp[i] / (mp[i][0] * 1.0 - (i == id))) + 1;
		}
	}
	if (dp[1] > 0) ans = min (ans, dp[1]);
}

int main () {
	scanf ("%d%d", &n, &m);
	rep (i, 1, m) {
		int x, y;
		scanf ("%d%d", &x, &y);
		node[i] = {x, y};
		mp[x][ ++mp[x][0] ] = y;
	}
	rep (i, 0, n) {
		calc (i);
	}
	printf ("%.7lf\n", ans);
	return 0;
}