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:
// 解答不能.
// codelegend氏.
// https://atcoder.jp/contests/abc144/submissions/8163936
#include <bits/stdc++.h>
using namespace std;
using LL = long long;
const int N = 600;
vector<int> adj[N];
double dp[N], dpr[N], prob[N];

int main(){
    int n, m;
    scanf("%d %d", &n, &m);
    for(int i = 0; i < m; i++){
        int u, v;
        scanf("%d %d", &u, &v);
        adj[u - 1].push_back(v - 1);
    }
    dp[n - 1] = 0;
    for(int u = n - 2; u >= 0; u--){
        double curr = 0;
        int sz = adj[u].size();
        for(int v: adj[u]) curr += dp[v];
        dp[u] = 1 + curr / sz;
        dpr[u] = dp[u];
        if(sz > 1) for(int v: adj[u]) dpr[u] = min(dpr[u], 1 + (curr - dp[v]) / (sz - 1));
    }
    prob[0] = 1;
    for(int u = 0; u < n - 1; u++){
        double pj = prob[u] / adj[u].size();
        for(int v: adj[u]) prob[v] += pj;
    }
    double ans = dp[0];
    for(int u = 0; u < n - 1; u++) ans = min(ans, dp[0] - prob[u] * (dp[u] - dpr[u]));
    printf("%.10lf", ans);
    return 0;
}