并查集,

删除节点操作,可以用新建节点代替

维护每个点到跟节点的距离

Bob has N (1 ≤ N ≤ 2*10⁵) gears (numbered from 1 to N). Each gear can rotate clockwise or counterclockwise. Bob thinks that assembling gears is much more exciting than just playing with a single one. Bob wants to put some gears into some groups. In each gear group, each gear has a specific rotation respectively, clockwise or counterclockwise, and as we all know, two gears can link together if and only if their rotations are different. At the beginning, each gear itself is a gear group.

Bob has M (1 ≤ N ≤ 4*10⁵) operations to his gears group:

• "L u v". Link gear u and gear v. If two gears link together, the gear groups which the two gears come from will also link together, and it makes a new gear group. The two gears will have different rotations. BTW, Bob won‘t link two gears with the same rotations together, such as linking (a, b), (b, c), and (a, c). Because it would shutdown the rotation of his gears group, he won‘t do that.
  
• "D u". Disconnect the gear u. Bob may feel something wrong about the gear. He will put the gear away, and the gear would become a new gear group itself and may be used again later. And the rest of gears in this group won‘t be separated apart.
  
• "Q u v". Query the rotations between two gears u and v. It could be "Different", the "Same" or "Unknown".
  
• "S u". Query the size of the gears, Bob wants to know how many gears there are in the gear group containing the gear u.

Since there are so many gears, Bob needs your help.

Input

Input will consist of multiple test cases. In each case, the first line consists of two integers N and M. Following M lines, each line consists of one of the operations which are described above. Please process to the end of input.

Output

For each query operation, you should output a line consist of the result.

Sample Input

3 7
L 1 2
L 2 3
Q 1 3
Q 2 3
D 2
Q 1 3
Q 2 3
5 10
L 1 2
L 2 3
L 4 5
Q 1 2
Q 1 3
Q 1 4
S 1
D 2
Q 2 3
S 1

Sample Output

Same
Different
Same
Unknown
Different
Same
Unknown
3
Unknown
2

Hint

Link (1, 2), (2, 3), (4, 5), gear 1 and gear 2 have different rotations, and gear 2 and gear 3 have different rotations, so we can know gear 1 and gear 3 have the same rotation, and we didn‘t link group (1, 2, 3) and group (4, 5), we don‘t know the situation about gear 1 and gear 4. Gear 1 is in the group (1, 2, 3), which has 3 gears. After putting gear 2 away, it may have a new rotation, and the group becomes (1, 3).

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

using namespace std;

const int maxn=600600;

int Sz[maxn],dist[maxn],fa[maxn],now[maxn],next;
int n,m;

void init(int n,int m)
{
    memset(dist,0,sizeof(dist));
    next=n+1;
    for(int i=1;i<=n+m;i++)
    {
        Sz[i]=1;fa[i]=i;
        if(i<=n) now[i]=i;
    }
}

int Find(int x)
{
    if(fa[x]==x) return x;
    int temp=Find(fa[x]);
    dist[x]+=dist[fa[x]];///到gen节点的距离
    return fa[x]=temp;
}

void Union(int x,int y)
{
    int X=Find(x),Y=Find(y);
    if(X==Y) return ;
    fa[X]=Y;
    Sz[Y]+=Sz[X];
    dist[X]=dist[Y]+1;
}

int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        init(n,m);
        char op[5];int a,b;
        while(m--)
        {
            scanf("%s",op);
            if(op[0]=='L')
            {
                scanf("%d%d",&a,&b);
                int na=now[a],nb=now[b];
                Union(na,nb);
            }
            else if(op[0]=='D')
            {
                scanf("%d",&a);
                int na=now[a];
                int A=Find(na);
                Sz[A]--;Sz[na]=0;
                now[a]=next++;
            }
            else if(op[0]=='S')
            {
                scanf("%d",&a);
                int na=now[a];
                int A=Find(na);
                printf("%d\n",Sz[A]);
            }
            else if(op[0]=='Q')
            {
                scanf("%d%d",&a,&b);
                int na=now[a],nb=now[b];
                int A=Find(na),B=Find(nb);
                if(A!=B) puts("Unknown");
                else
                {
                    if ((dist[na]-dist[nb])&1) puts("Different");
                    else puts("Same");
                }
            }
        }
    }
    return 0;
}