During the War of Resistance Against Japan, tunnel warfare was carried out extensively in the vast areas of north China Plain. Generally speaking, villages connected by tunnels lay in a line. Except the two at the ends, every village was directly connected with two neighboring ones.

Frequently the invaders launched attack on some of the villages and destroyed the parts of tunnels in them. The Eighth Route Army commanders requested the latest connection state of the tunnels and villages. If some villages are severely isolated, restoration of connection must be done immediately!

The first line of the input contains two positive integers n and m (n, m ≤ 50,000) indicating the number of villages and events. Each of the next m lines describes an event.

There are three different events described in different format shown below:

1. D x: The x-th village was destroyed.
2. Q x: The Army commands requested the number of villages that x-th village was directly or indirectly connected with including itself.
3. R: The village destroyed last was rebuilt.

Output the answer to each of the Army commanders’ request in order on a separate line.

An illustration of the sample input:

      OOOOOOOD 3   OOXOOOOD 6   OOXOOXOD 5   OOXOXXOR     OOXOOXOR     OOXOOOO

#include<iostream>#include<algorithm>#include<cstdlib>#include<sstream>#include<cstring>#include<bitset>#include<cstdio>#include<string>#include<deque>#include<stack>#include<cmath>#include<queue>#include<set>#include<map>using namespace std;#define INF 0x3f3f3f3f#define CLR(x,y) memset(x,y,sizeof(x))#define LC(x) (x<<1)#define RC(x) ((x<<1)+1)#define MID(x,y) ((x+y)>>1)typedef pair<int,int> pii;typedef long long LL;const double PI=acos(-1.0);const int N=50010;struct seg{    int l,mid,r;    int L,R,M;    int len()    {        return r-l+1;    }};seg T[N<<2];void pushup(int k){    T[k].L=T[LC(k)].L;//暂时保持    T[k].R=T[RC(k)].R;//暂时保持    if(T[k].L==T[LC(k)].len())//可以向右拓展        T[k].L+=T[RC(k)].L;//加上向右可拓展的长度    if(T[k].R==T[RC(k)].len())//可以向左拓展        T[k].R+=T[LC(k)].R;//加上向左可拓展的长度    T[k].M=max<int>(T[LC(k)].R+T[RC(k)].L,max<int>(T[LC(k)].M,T[RC(k)].M));//更新M}void build(int k,int l,int r){    T[k].l=l;    T[k].r=r;    T[k].mid=MID(l,r);    if(l==r)    {        T[k].L=1;        T[k].R=1;        T[k].M=1;        return ;    }    build(LC(k),l,T[k].mid);    build(RC(k),T[k].mid+1,r);    pushup(k);//记得pushup}void update(int k,int x,int val){    if(T[k].l==T[k].r&&T[k].l==x)    {        T[k].M=T[k].L=T[k].R=val;        return ;    }    if(x<=T[k].mid)        update(LC(k),x,val);    else        update(RC(k),x,val);    pushup(k);}int query(int k,int x){    if(T[k].l==T[k].r||T[k].M==0||T[k].len()==T[k].M)        return T[k].M;    else    {        if(x<=T[k].mid)//在左子树        {            if(x>=T[k].mid-T[LC(k)].R+1)//若在左子树的R范围内则说明可以继续向右子树连通                return query(LC(k),x)+query(RC(k),T[RC(k)].l);            else                return query(LC(k),x);        }        else//在右子树        {            if(x<=(T[k].mid+1)+T[RC(k)].L-1)//若在右子树的L范围内说明可以继续向左子树连通                return query(RC(k),x)+query(LC(k),T[LC(k)].r);            else                return query(RC(k),x);        }    }}int main(void){    int n,m,x,r;    char ops[4];    while (~scanf("%d%d",&n,&m))    {        build(1,1,n);        stack<int>Q;        while (m--)        {            scanf("%s%d",ops,&x);            if(ops[0]==‘D‘)            {                update(1,x,0);                Q.push(x);            }            else if(ops[0]==‘R‘)            {                update(1,Q.top(),1);                Q.pop();            }            else                printf("%d\n",query(1,x));        }    }    return 0;}