Problem: Let us compare two triples a = (xa, ya, za) and b = (xb, yb, zb) by a partial order ∠ defined as follows.
a ∠ b ⇔ xa < xb and ya < yb and za < zb



Your mission is to find, in the given set of triples, the longest ascending series a1 ∠ a2 ∠ ... ∠ ak.



Input

The input is a sequence of datasets, each specifying a set of triples formatted as follows.


m n A B
x1 y1 z1
x2 y2 z2
...
xm ym zm


Here, m, n, A and B in the first line, and all of xi, yi and zi (i = 1, . . . , m) in the following lines are non-negative integers.

Each dataset specifies a set of m + n triples. The triples p1 through pm are explicitly specified in the dataset, the i-th triple pi being (xi, yi, zi). The remaining n triples are specified by parameters A and B given to the following generator routine.


int a = A, b = B, C = ~(1<<31), M = (1<<16)-1;
int r() {
a = 36969 * (a & M) + (a >> 16);
b = 18000 * (b & M) + (b >> 16);
return (C & ((a << 16) + b)) % 1000000;
}


Repeated 3n calls of r() defined as above yield values of xm+1, ym+1, zm+1, xm+2, ym+2, zm+2, ..., xm+n, ym+n, and zm+n, in this order.

You can assume that 1 ≤ m + n ≤ 3 × 105, 1 ≤ A,B ≤ 216, and 0 ≤ xk, yk, zk < 106 for 1 ≤ k ≤ m + n.

The input ends with a line containing four zeros. The total of m + n for all the datasets does not exceed 2 × 106.

Output

For each dataset, output the length of the longest ascending series of triples in the specified set. If pi1 ∠ pi2 ∠ ... ∠ pik is the longest, the answer should be k.

Example

Input

6 0 1 1
0 0 0
0 2 2
1 1 1
2 0 2
2 2 0
2 2 2
5 0 1 1
0 0 0
1 1 1
2 2 2
3 3 3
4 4 4
10 0 1 1
3 0 0
2 1 0
2 0 1
1 2 0
1 1 1
1 0 2
0 3 0
0 2 1
0 1 2
0 0 3
0 10 1 1
0 0 0 0


Output

3
5
1
3

Solution (in CPP):

Solution:
#include<cstdio>
#include<set>
#include<vector>
#include<utility>
#include<algorithm>

using namespace std;

typedef pair<int,int> P;
typedef pair<int,P> PP;

typedef set<P>::iterator se_it;

set<P> se[300300];

int xs[300300],ys[300300],zs[300300];
PP pps[300300];
int N;

bool check(P p,int id){
	set<P> &se=::se[id];
	int x=p.first,y=p.second;
	if(se.size()==0) return false;
	se_it it=se.lower_bound(P(x,-2));
	if(it==se.begin()) return false;
	it--;
	int py=it->second;
	if(py<y) return true;
	return false;
}

void ins(P p,int id){
	set<P> &se=::se[id];
	int x=p.first,y=p.second;
	se_it it=se.lower_bound(P(x,-2));
	if(it==se.end()){
		se.insert(p);
		return;
	}
	if(it->first==x&&it->second<y){
		return;
	}
	while(true){
		if(it==se.end()||it->second<y) break;
		se.erase(it++);
	}
	se.insert(p);
}

void put(P p){
	int lb=0,ub=N+1;
	while(ub-lb>1){
		int mid=(ub+lb)/2;
		bool flg=check(p,mid);
		if(flg) lb=mid;
		else ub=mid;
	}
	ins(p,lb+1);
}

int A,B;
int a, b, C = ~(1<<31), M = (1<<16)-1;
int r() {
  a = 36969 * (a & M) + (a >> 16);
  b = 18000 * (b & M) + (b >> 16);
  return (C & ((a << 16) + b)) % 1000000;
}

int m,n;

void init(){
	scanf("%d%d%d%d",&m,&n,&A,&B);
	if(m+n==0) exit(0);
	for(int i=0;i<m;i++){
		scanf("%d%d%d",xs+i,ys+i,zs+i);
	}
	a=A,b=B;
	for(int i=0;i<n;i++){
		xs[i+m]=r();
		ys[i+m]=r();
		zs[i+m]=r();
	}
	N=m+n;
	for(int i=0;i<=N;i++){
		se[i].clear();
	}
	for(int i=0;i<N;i++){
		pps[i]=PP(zs[i],P(-xs[i],-ys[i]));
	}
	sort(pps,pps+N);
}

int solve(){
	se[0].insert(P(-1,-1));
	for(int i=0;i<N;i++){
		P p=pps[i].second;
		p.first*=-1;
		p.second*=-1;
		put(p);
	}
	int ans=0;
	for(int i=0;i<=N;i++) if(se[i].size()>0) ans=i;
	return ans;
}

int main(){
	while(true){
		init();
		int ans=solve();
		printf("%d\n",ans);
	}
	return 0;
}