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;
#define what_is(x) cerr << #x << " is " << x << endl;
#define IOS ios::sync_with_stdio(false); cin.tie(0);
#define st first
#define nd second
#define endl '\n'

typedef long long ll;
typedef pair<int,int> pii;
const int N = 600+5;
const int INF = 1e9;
const int MOD = 1e9+7;

int n, m;
vector<int> g[N];
double ans = 1e9;

void block(int v){
	vector<double> dp(n+2);
	dp[n] = 0.0;
	for(int i=n-1; i>=1; i--){
		int sz = g[i].size();
		double sum = 0.0;
		double mx = 0.0;
		for(auto u:g[i]){
			sum += dp[u];
			mx = max(mx, dp[u]);
		}
		if (i == v && sz > 1){
			sum -= mx;
			sz -= 1;
		}
		dp[i] = sum / sz + 1; //1 for all edges
	}
	ans = min(ans, dp[1]);
}

int main()
{
	IOS
	// freopen("input.txt", "r", stdin);
	cin >> n >> m;
	for(int i=0; i<m; i++){
		int u, v; cin >> u >> v;
		g[u].push_back(v);
	}
	for(int i=0; i<n; i++){ //i=0 means no block any passage
		block(i);
	}
	cout << fixed << setprecision(9);
	cout << ans << endl;
	return 0;
}