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;
vector<int> G1[605];
vector<int> G2[605];
double dp1[605];
double dp2[605];
int main()
{
    int n, m;
    scanf("%d %d", &n, &m);
    for(int i = 0; i < m; i++) {
        int s, t;
        scanf("%d %d", &s, &t);
        G1[s].push_back(t);
        G2[t].push_back(s);
    }
    dp1[1] = 1.0;
    for(int i = 1; i < n; i++) {
        int size = G1[i].size();
        for(int nxt : G1[i]) {
            dp1[nxt] += dp1[i] / size;
        }
    }
    for(int i = n; i > 0; i--) {
        int size = G2[i].size();
        for(int nxt : G2[i]) {
            dp2[nxt] += (dp2[i] + 1.0) / G1[nxt].size();
        }
    }
    double ans = dp2[1];
    for(int i = 1; i < n; i++) {
        if(G1[i].size() <= 1) continue;
        int size = G1[i].size();
        double ma = 0, sum = 0;
        double rate = dp1[i] / size;
        for(int nxt : G1[i]) {
            sum += dp2[nxt];
            ma = max(dp2[nxt], ma);
        }
        ans = min(ans, dp2[1] - sum * rate + (sum - ma) * dp1[i] / (size - 1));
    }
    printf("%.10f\n", ans);
    return 0;
}