cf-707D

/* In this problem persistent data structure is needed..It means we have to keep a track of the previous updates.Because in some query it is said to restore the form after the kth update...It can be solved using a simple dfs..Where we make edges like ....

after performing the ith query we make an edge to i to i+1 th query..whenever it is said to restore the kth version then we make an edge to k to ith query....Remember when dfs traversal is finished in a node we have to discard all changes made due to this query...Then simply proceed to next node....*/


vector<piii>v[MAX];int ans[MAX];

bitset<1002>b[1002],bit;int tot;

void dfs(int s)
{
    for(auto it : v[s])
    {
        int type = it.ff.ff;
        int pos = it.ff.ss;
        int r = it.ss.ff;
        int c = it.ss.ss;
        if(type==1)
        {
            int val = b[r][c];
            tot-=b[r].count();
            b[r][c] = 1;
            tot+=b[r].count();
            ans[pos] = tot;
            dfs(pos);
            tot-=b[r].count();
            b[r][c] = val;
            tot+=b[r].count();

        }
        if(type==2)
        {
            int val = b[r][c];
            tot-=b[r].count();
            b[r][c] = 0;
            tot+=b[r].count();
            ans[pos] = tot;
            dfs(pos);
            tot-=b[r].count();
            b[r][c] = val;
            tot+=b[r].count();

        }
        if(type==3)
        {
            tot-=b[r].count();
            b[r]^=bit;
            tot+=b[r].count();
            ans[pos] = tot;
            dfs(pos);
            tot -= b[r].count();
            b[r]^=bit;
            tot += b[r].count();
        }
        if(type==4)
        {
            ans[pos] = tot;
            dfs(pos);
        }
    }
}


int main()
{
    booster()
    ///read("input.txt");

    int n,m,q;
    cin>>n>>m>>q;
    for(int i = 1;i<=m;i++)bit[i] = 1;
    int cur = 0;
    for(int i = 1;i<=q;i++)
    {
        int c;
        int x,y;
        cin>>c>>x;
        if(c==1||c==2)cin>>y;
        if(c!=4)
        {
            v[cur].pb(piii(pii(c,i),pii(x,y)));
        }
        else
        {
            v[x].pb(piii(pii(c,i),pii(x,y)));
        }
        cur = i;

    }

    dfs(0);

    for(int i = 1;i<=q;i++)cout<<ans[i]<<endl;

    return 0;
}