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<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;

const int N = 605;
const double INF = 1e14;

int n, m, u, v;

struct Edge{
	int nxt, to;
}edge[N*N];
int head[N], edgeNum, deg[N];
void addEdge(int from, int to){
	edge[++edgeNum] = (Edge){head[from], to};
	head[from] = edgeNum;
	deg[from]++;
}

double dp[N];
int tdeg[N];
double blockDP(int x){
	for(int i = 1; i <= n; i++){
		dp[i] = INF;
		tdeg[i] = deg[i];
	}
	dp[n] = 0;
	for(int i = n - 1; i >= 1; i--){
		double sumDP = 0, maxDP = 0;
		for(int k = head[i]; k; k = edge[k].nxt){
			maxDP = max(maxDP, dp[edge[k].to] == INF ? 0 : dp[edge[k].to]);
			sumDP += dp[edge[k].to];
		}
		if(i == x && tdeg[i] >= 2){
			sumDP -= maxDP;
			tdeg[i]--;
		}
		dp[i] = sumDP / tdeg[i] + 1;
	}
	return dp[1];
}

int main(){
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= m; i++){
		scanf("%d%d", &u, &v);
		addEdge(u, v);
	}
	double ans = INF;
	for(int i = 1; i <= n; i++)
		ans = min(ans, blockDP(i));
	printf("%.10f\n", ans);
	return 0;
}