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:
//差分計算を頑張ろうとすると沼る。冷静になると、消す辺の候補を高々N本に絞れる。
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstdio>
#define rep(i, n) for(i = 0; i < n; i++)
using namespace std;

int n, m;
vector<int> et[600];
double E[600];

int main() {
	int i, j, k;
	
	cin >> n >> m;
	rep(i, m) {
		int s, t;
		cin >> s >> t; s--; t--;
		et[s].push_back(t);
	}
	
	for (i = n - 2; i >= 0; i--) {
		rep(j, et[i].size()) {
			E[i] += E[et[i][j]];
		}
		E[i] /= et[i].size();
		E[i] += 1;
	}

	double ans = 11451419;
	rep(i, n) {
		double maxE = -1;
		int maxJ = -1;
		rep(j, et[i].size()) {
			if (maxE < E[et[i][j]]) {
				maxE = E[et[i][j]];
				maxJ = j;
			}
		}
		
		double F[600] = {0};
		for (j = n - 2; j >= 0; j--) {
			int cnt = 0;
			rep(k, et[j].size()) {
				if (j == i && k == maxJ) continue;
				F[j] += F[et[j][k]];
				cnt++;
			}
			if (cnt == 0) break;
			F[j] /= cnt;
			F[j] += 1;
		}
		if (j == -1) ans = min(ans, F[0]);
	}
	printf("%.14f\n", ans);
	return 0;
}