An earthquake takes place in Southeast Asia. The ACM (Asia Cooperated Medical team) have set up a wireless network with the lap computers, but an unexpected aftershock attacked, all computers in the network were all broken. The computers are repaired one by one, and the network gradually began to work again. Because of the hardware restricts, each computer can only directly communicate with the computers that are not farther than d meters from it. But every computer can be regarded as the intermediary of the communication between two other computers, that is to say computer A and computer B can communicate if computer A and computer B can communicate directly or there is a computer C that can communicate with both A and B.

In the process of repairing the network, workers can take two kinds of operations at every moment, repairing a computer, or testing if two computers can communicate. Your job is to answer all the testing operations.

Input

The first line contains two integers N and d (1 <= N <= 1001, 0 <= d <= 20000). Here N is the number of computers, which are numbered from 1 to N, and D is the maximum distance two computers can communicate directly. In the next N lines, each contains two integers xi, yi (0 <= xi, yi <= 10000), which is the coordinate of N computers. From the (N+1)-th line to the end of input, there are operations, which are carried out one by one. Each line contains an operation in one of following two formats:
1. "O p" (1 <= p <= N), which means repairing computer p.
2. "S p q" (1 <= p, q <= N), which means testing whether computer p and q can communicate.

The input will not exceed 300000 lines.

Output

For each Testing operation, print "SUCCESS" if the two computers can communicate, or "FAIL" if not.

Sample Input

4 1
0 1
0 2
0 3
0 4
O 1
O 2
O 4
S 1 4
O 3
S 1 4

Sample Output

FAIL
SUCCESS

题意：

      电脑被全部损坏，给出各个电脑的坐标，修好的电脑之间距离不超过d可以联系，也可以通过其他修好的电脑间接相连。‘O’表示要维修的电脑，‘S’后有两个询问的电脑x , y，询问他们是否能够联系。

思路：

     判断两台修好的电脑距离与d之间的关系，如果小于等于d就合并，最后判断询问的x,y是否是同一个祖先。

    读入字符的的时候要小心，一开始忘记了！=EOF结果时间超限好几次。

代码：

#include<stdio.h>
#include<string.h>
int f[1010],n,d,book[1010];
int x[1010],y[1010];
void init()
{
	int i;
	for(i=1;i<=n;i++)	
		f[i]=i;
	return;
}
int dis(int i,int j)
{
	if((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j])<=d*d)
		return 1;
	return 0;
}
int getf(int v)
{
	if(f[v]==v)
		return v;
	else
		f[v]=getf(f[v]);
	return f[v];
}
void merge(int x,int y)
{
	int t1,t2;
	t1=getf(x);
	t2=getf(y);
	if(t1!=t2)
	{
		f[t2]=t1;
	}
	return;
}
int main()
{
	int i,j,k,a,b;
	char ch;
	while(scanf("%d%d",&n,&d)!=EOF)
	{
		init();
		memset(book,0,sizeof(book));
		for(i=1;i<=n;i++)
			scanf("%d%d",&x[i],&y[i]);
		while(scanf("%c",&ch)!=EOF)
		{
			if(ch=='O')
			{
				scanf("%d",&a);
				book[a]=1;
				for(i=1;i<=n;i++)
				{
					if(i!=a&&book[i]==1&&dis(i,a))
					{
						merge(i,a);
					}
				}
			}
			if(ch=='S')
			{
				scanf("%d%d",&a,&b);
				if(getf(a)==getf(b))
					printf("SUCCESS\n");
				else
					printf("FAIL\n");
			}
		}
	}
	return 0;
}