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 in std::cin
#define out std::cout
#define rep(i,N) for(LL i=0;i<N;++i)
typedef long long int LL;
typedef long double LD;

const LL inf = 1123456789;
std::vector<std::vector<LL>>G;
std::vector<LD>memo;
LL N, unable;

LD dp(LL v)
{
	if (v == N) return 0;
	if (memo[v] != -1) return memo[v];
	LD sum = 0, max = 0, res;
	for (LL u : G[v])
	{
		sum += dp(u) + 1;
		max = std::max(max, dp(u) + 1);
	}
	if (v == unable && G[v].size() >= 2) res = (sum - max) / (G[v].size() - 1);
	else res = sum / G[v].size();
	return memo[v] = res;
}

int main()
{
	LL M;
	in >> N >> M;
	std::vector<LL>s(M), t(M);
	rep(i, M) in >> s[i] >> t[i];

	G.resize(N + 1);
	rep(i, M) G[s[i]].push_back(t[i]);

	LD ans = inf;
	for (unable = 1; unable <= N; ++unable)
	{
		memo.clear();
		memo.resize(N + 1, -1.);
		ans = std::min(ans, dp(1));
	}

	out << std::fixed << std::setprecision(10);
	out << ans << std::endl;
}