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:
// F.cpp
#include <bits/stdc++.h>

using namespace std;

const int MAX_N = 660;

int head[MAX_N], current, n, m;
double dp[MAX_N];
bool vis[MAX_N];

struct edge
{
    int to, nxt;
    bool tag;
} edges[MAX_N * MAX_N << 1];

void addpath(int src, int dst)
{
    edges[current].to = dst, edges[current].nxt = head[src];
    head[src] = current++;
}

void process(int block)
{
    for (int u = n - 1; u >= 1; u--)
    {
        int deg = 0;
        double max_son = 0, sum = 0;
        for (int i = head[u]; i != -1; i = edges[i].nxt)
            sum += dp[edges[i].to], max_son = max(max_son, dp[edges[i].to]), deg++;
        if (block == u && deg >= 2)
            deg--, sum -= max_son;
        dp[u] = sum / deg + 1;
    }
}

int main()
{
    memset(head, -1, sizeof(head));
    scanf("%d%d", &n, &m);
    for (int i = 1, u, v; i <= m; i++)
        scanf("%d%d", &u, &v), addpath(u, v);
    double ans = 0x3f3f3f3f;
    for (int i = 0; i <= n - 1; i++)
        process(i), ans = min(ans, dp[1]);
    printf("%.8lf", ans);
    return 0;
}