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>

using namespace std;

#ifdef NeverBeRed
#include "debug.h"
#else
#define debug(...) 9715
#endif
typedef long long ll;
typedef long double ld;
typedef complex<ll> point;
#define F first
#define S second

ld dp[2][605];

int main()
{
	#ifdef TurnRed
		freopen("a.in", "r", stdin);
		//freopen("a.out", "w", stdout);
	#endif

	ios_base::sync_with_stdio(0), cin.tie(0);

	int n, m;
	cin >> n >> m;
	vector<vector<int>> adj(n);
	for (int u, v; m--; )
	{
		cin >> u >> v;
		--u, --v;
		adj[u].push_back(v);
	}

	ld ans = 1e60;
	for (int i = 0; i < n; ++i)
	{
		memset(dp, 0, sizeof dp);
		for (int u = n-2; u >= 0; --u)
		{
			int sz = adj[u].size();
			for (auto v : adj[u]) dp[0][u] += dp[v==i][v];
			dp[0][u] = dp[0][u] / sz + 1;
			dp[1][u] = dp[0][u];
			if (sz > 1 && u == i)
			{
				ld s = 0;
				for (auto v : adj[u])
					s += dp[0][v];
				for (auto v : adj[u])
					dp[1][u] = min(dp[1][u], (s - dp[0][v]) / (sz-1) + 1);
			}
		}
		ans = min(ans, dp[1][0]);
	}
	cout << setprecision(20) << fixed << ans << "\n";

	return 0;
}