/*
Vom considera V[i] = 1 : camera libera si V[i] = -1 : camera ocupata
Vom lucra cu arbori de intervale asemanator problemei : subsecventa de suma maxima intervalului [l, r]
*/

#include <fstream>

using namespace std;

ifstream f("hotel.in");
ofstream g("hotel.out");

const int NMAX = 100002;

int N, P;
int Sum[4 * NMAX];
int Left[4 * NMAX];
int Right[4 * NMAX];
int Best[4 * NMAX];
int A, B, val;

void Build(int nod, int st, int dr)
{
    if (st == dr)
    {
        Sum[nod] = Left[nod] = Right[nod] = Best[nod] = 1;
        return;
    }
    int mid = (st + dr) >> 1;
    Build (nod << 1, st, mid);
    Build ((nod << 1) + 1, mid + 1, dr);
    Sum[nod] = Left[nod] = Right[nod] = Best[nod] = Best[nod << 1] + Best[(nod << 1) + 1];
}
void Update(int nod, int st, int dr)
{
    if (A <= st && dr <= B)
    {
        Sum[nod] = Left[nod] = Right[nod] = Best[nod] = (dr - st + 1) * val;
        return;
    }

    int mid = (st + dr) >> 1;

    if (Best[nod] == (dr - st + 1) * (-1))
    {
        Sum[nod << 1] = Left[nod << 1] = Right[nod << 1] = Best[nod << 1] =  (mid - st + 1) * (-1);
        Sum[(nod << 1) + 1] = Left[(nod << 1) + 1] = Right[(nod << 1) + 1] = Best[(nod << 1) + 1] = (dr - mid) * (-1);
    }
    else if (Best[nod] == dr - st + 1)
    {
        Left[nod << 1] = Right[nod << 1] = Best[nod << 1] = mid - st + 1;
        Left[(nod << 1) + 1] = Right[(nod << 1) + 1] = Best[(nod << 1) + 1] = dr - mid;
    }

    if (A <= mid)
        Update(nod << 1, st, mid);
    if (B > mid)
        Update((nod << 1) + 1, mid + 1, dr);

    Sum[nod] = Sum[(nod << 1)] + Sum[(nod << 1) + 1];
    Left[nod] = max(Left[(nod << 1)], Sum[(nod << 1)] + Left[(nod << 1) + 1]);
    Right[nod] = max(Right[(nod << 1) + 1], Sum[(nod << 1) + 1] + Right[(nod << 1)]);
    Best[nod] = max(max(Best[(nod << 1)], Best[(nod << 1) + 1]), Left[(nod << 1) + 1] + Right[(nod << 1)]);
}

int Query()
{
    return Best[1];
}


int main()
{
    f >> N >> P;
    Build (1, 1, N);
    for (int i = 1; i <= P; ++i)
    {
        int tip;
        f >> tip;
        if (tip == 3)
            g << Query() << "\n";
        else
        {
            f >> A >> B;
            B += A - 1;
            if (tip == 1)
                val = -1;
            else
                val = 1;
            Update (1, 1, N);
        }
    }
    f.close();
    g.close();
    return 0;
}