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:

      OOOOOOO

D 3   OOXOOOO

D 6   OOXOOXO

D 5   OOXOXXO

R     OOXOOXO

R     OOXOOOO

#include <stdio.h>

#include <iostream>

#include <algorithm>

#include <cstdlib>

#include <sstream>

#include <numeric>

#include <cstring>

#include <bitset>

#include <string>

#include <deque>

#include <stack>

#include <cmath>

#include <queue>

#include <set>

#include <map>

using namespace std;

#define INF 0x3f3f3f3f

#define LC(x) (x<<1)

#define RC(x) ((x<<1)+1)

#define MID(x,y) ((x+y)>>1)

#define CLR(arr,val) memset(arr,val,sizeof(arr))

#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);

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,M,R;

    int len;

};

seg T[N<<2];

int st[N],top;

inline void pushup(int k)

{

    T[k].L=T[LC(k)].L;

    T[k].R=T[RC(k)].R;

    if(T[k].L==T[LC(k)].len)

        T[k].L+=T[RC(k)].L;

    if(T[k].R==T[RC(k)].len)

        T[k].R+=T[LC(k)].R;

    T[k].M=max<int>(T[LC(k)].R+T[RC(k)].L,max<int>(T[LC(k)].M,T[RC(k)].M));

}

void build(int k,int l,int r)

{

    T[k].l=l;

    T[k].r=r;

    T[k].mid=MID(l,r);

    T[k].len=r-l+1;

    T[k].L=T[k].R=T[k].M=r-l+1;

    if(l==r)

        return ;

    build(LC(k),l,T[k].mid);

    build(RC(k),T[k].mid+1,r);

}

void update(int k,int x,char flag)

{

    if(T[k].l==T[k].r)

    {

        if(flag=='D')

            T[k].L=T[k].R=T[k].M=0;

        else

            T[k].L=T[k].R=T[k].M=1;

    }

    else

    {

        if(x<=T[k].mid)

            update(LC(k),x,flag);

        else

            update(RC(k),x,flag);

        pushup(k);

    }

}

int query(int k,int x)

{

    if(T[k].M==T[k].len||!T[k].M||T[k].len==1)

        return T[k].M;

    else

    {

        if(x<=T[k].mid)

        {

            if(x>=T[k].mid+1-T[LC(k)].R)

                return query(LC(k),x)+query(RC(k),T[k].mid+1);

            else

                return query(LC(k),x);

        }

        else

        {

            if(x<=T[k].mid+T[RC(k)].L)

                return query(LC(k),T[k].mid)+query(RC(k),x);

            else

                return query(RC(k),x);

        }

    }

}

int main(void)

{

    char ops[3];

    int n,m,x;

    while (~scanf("%d%d",&n,&m))

    {

        top=0;

        build(1,1,n);

        while (m--)

        {

            scanf("%s",ops);

            if(ops[0]=='D')

            {

                scanf("%d",&x);

                update(1,st[top++]=x,'D');

            }

            else if(ops[0]=='Q')

            {

                scanf("%d",&x);

                printf("%d\n",query(1,x));

            }

            else

                update(1,st[--top],'R');

        }

    }

    return 0;

}