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 JAVA):

Solution:
import java.util.*;
import java.util.stream.Collectors;

public class Main {

    void run() {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();

        List<Integer>[] lists = new List[n];

        for (int i = 0; i < n; i++) {
            lists[i] = new ArrayList<>();
        }

        for (int i = 0; i < m; i++) {
            int s = sc.nextInt() - 1;
            int t = sc.nextInt() - 1;
            lists[s].add(t);
        }

        double[] baseDP = new double[n];
        int[] maxCostNodes = new int[n];

        for (int i = n - 2; i >= 0; i--) {
            double max = -1;
            int maxCostNode = -1;
            List<Integer> list = lists[i];
            for (int node : list) {
                if (max < baseDP[node]) {
                    max = baseDP[node];
                    maxCostNode = node;
                }
                baseDP[i] += baseDP[node];
            }
            baseDP[i] = baseDP[i] / list.size() + 1;
            maxCostNodes[i] = maxCostNode;
        }

//        debug(baseDP);
//        debug(maxCostNodes);

        double ans = m;

        for (int i = 0; i < n; i++) {
            double[] dp = new double[n];
            final int maxCostNode = maxCostNodes[i];

            for (int j = n - 2; j >= 0; j--) {
                List<Integer> list = i == j && lists[j].size() > 1
                        ? lists[j].stream().filter(x -> x != maxCostNode).collect(Collectors.toList())
                        : lists[j];
                for (int node : list) {
                    dp[j] += dp[node];
                }
                dp[j] = dp[j] / list.size() + 1;
            }
//            debug(dp);

            ans = Math.min(dp[0], ans);
        }

        System.out.println(ans);

    }

    void debug(Object... os) {
        System.err.println(Arrays.deepToString(os));
    }

    public static void main(String[] args) {
        new Main().run();
    }
}