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 <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <map>
#include <unordered_map>
#include <set>
#include <functional>
#include <bitset>
#include <cassert>
#include <random>
using namespace std;
using ll = long long;


int main() {
	int n, m; cin >> n >> m;
	vector<vector<int>> edg(n), eid(n);
	for (int i = 0; i < m; i++) {
		int u, v; cin >> u >> v;
		u--; v--;
		edg[u].push_back(v);
		eid[u].push_back(i);
	}
	double ans = 1e18;
	vector<double> dp(n);
	for (int i = -1; i < n; i++) {
		for (int j = 0; j < n; j++) dp[j] = 0.;
		for (int j = n - 2; j >= 0; j--) {
			int c = edg[j].size();
			double ma = 0.;
			for (int t = 0; t < c; t++) {
				dp[j] += dp[edg[j][t]];
				if (edg[j][t] != n - 1) {
					ma = max(ma, dp[edg[j][t]]);
				}
			}
			if (j == i && c > 1) {
				dp[j] -= ma;
				dp[j] /= (c - 1);
			}
			else {
				dp[j] /= c;
			}
			dp[j]++;
		}
		ans = min(ans, dp[0]);
	}
	printf("%.15f\n", ans);
	return 0;
}