Description

The police office in Tadu City decides to say ends to the chaos, as launch actions to root up the TWO gangs in the city, Gang Dragon and Gang Snake. However, the police first needs to identify which gang a criminal belongs to. The present question is, given two criminals; do they belong to a same clan? You must give your judgment based on incomplete information. (Since the gangsters are always acting secretly.)

Assume N (N <= 10^5) criminals are currently in Tadu City, numbered from 1 to N. And of course, at least one of them belongs to Gang Dragon, and the same for Gang Snake. You will be given M (M <= 10^5) messages in sequence, which are in the following two kinds:

1. D [a] [b]

where [a] and [b] are the numbers of two criminals, and they belong to different gangs.

2. A [a] [b]

where [a] and [b] are the numbers of two criminals. This requires you to decide whether a and b belong to a same gang.

Input

The first line of the input contains a single integer T (1 <= T <= 20), the number of test cases. Then T cases follow. Each test case begins with a line with two integers N and M, followed by M lines each containing one message as described above.

Output

For each message "A [a] [b]" in each case, your program should give the judgment based on the information got before. The answers might be one of "In the same gang.", "In different gangs." and "Not sure yet."

Sample Input

15 5A 1 2D 1 2A 1 2D 2 4A 1 4

Sample Output

Not sure yet.In different gangs.In the same gang.

1 #include
       
         2 #include
        
          3 int rank[1000000],pa[1000000]; 4 int cha(int k) 5 { 6     if(pa[k]!=k) 7     { 8         int t=pa[k]; 9         pa[k]=cha(pa[k]);10         rank[k]=(rank[k]+rank[t])%2;//不断更新11     }12     return pa[k];13 }14 bool bing(int x,int y)15 {16     int x2=cha(x);17     int y2=cha(y);18     if(x2!=y2)19     {20         pa[y2]=x2;//设为同一根节点后（不用根节点区分），方便用rank的值来区分21         rank[y2]=(rank[x]-rank[y]+1)%2;//用于标记它俩是否是属于同一帮派22     }23     return true;24 }25 void init(int n)26 {27     for(int i=1; i<=n; i++)28     {29         pa[i]=i;30         rank[i]=0;31     }32 }33 int main()34 {35     int cas,i,j,m,n,p1,p2,x2,y2;36     char k[2];37     scanf("%d",&cas);38     while(cas--)39     {40 41         scanf("%d%d",&n,&m);//n代表人数，m代表测试数据42         init(n);43         for(i=1; i<=m; i++)44         {45             scanf("%s%d%d",k,&p1,&p2);46             if(k[0]=='A')47             {48                 x2=cha(p1);49                 y2=cha(p2);50                 if(x2!=y2)51                     printf("Not sure yet.\n");//是第一次出现，根节点不相同时，还不能确定是否为同一帮派52                 else53                 {54                     if(rank[p2]==rank[p1])55                         printf("In the same gang.\n");56                     else57                         printf("In different gangs.\n");58                 }59             }60             if(k[0]=='D')61             {62                 bing(p1,p2);63             }64         }65 66     }67     return 0;68 }
        
       

2015-07-03，15:45:47