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;
using ll  = long long;
using vl  = vector<ll>;
using vvl = vector<vl>;
using ld  = long double;
using vd  = vector<ld>;
#define rng(i,a,b) for(int i = (a); i < (b); ++i)
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define per(i,n) for(int i = (n)-1; i >= 0; --i)

int N, M;
vvl edges;
ld ans;

int main() {
    cin >> N >> M;
    edges.resize(N);
    ans = 1e9;
    rep(i,M) {
        int a, b; cin >> a >> b; a--; b--;
        edges[a].push_back(b);
    }
    rep(k,N) {
        vd dp(N, 0);
        per(i,N) {
            ld sum = 0, mx = 0;
            int sz = edges[i].size();
            for (auto j : edges[i]) {
                mx = max(mx, dp[j]);
                sum += dp[j];
            }
            if (i == k and sz >= 2) { sum -= mx; sz--; }
            if (sz >= 1) sum = (sum + sz) / sz;
            dp[i] = sum;
        }
        ans = min(ans, dp[0]);
    }
    printf("%.10Lf\n", ans);
    return 0;
}