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 <bits/stdc++.h>
using namespace std;

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;
}

class HappySequence{
public:
	map<int,int> seq;
	HappySequence(){ int inf = 1e9; seq[-inf] = -inf; }
	bool insert(int key,int val){
		if( (--seq.lower_bound(key))->second >= val ) return false;
		seq[key] = val;
		auto it = ++seq.find(key);
		while( it != seq.end() && (*it).second <= val ) seq.erase(it++);
		return true;
	}
};

HappySequence seg[1000100];

void ins(int y,int z,int v){
	for(int i = y ; i < 1000100 ; i += i & -i ) seg[i].insert(z,v);
}
int get(int y,int z){
	int v = 0;
	for(int i = y ; i > 0 ; i -= i & -i ){
		v = max(v,(--seg[i].seq.lower_bound(z))->second);
	}
	return v;
}

int main(){
	int n,m,A,B;
	while( cin >> m >> n >> A >> B && n+m ){	
		a = A;
		b = B;
		vector< array<int,3> > ps;
		vector<int> uy;
		for(int i = 0 ; i < m ; i++){	
			int x,y,z;
			cin >> x >> y >> z;
			ps.push_back(array<int,3>{x,y,z});
			uy.push_back(y);
		}
		for(int i = 0 ; i < n ; i++){
			int x = r();
			int y = r();
			int z = r();
			ps.push_back(array<int,3>{x,y,z});
			uy.push_back(y);
		}
		sort(uy.begin(),uy.end());
		uy.erase(unique(uy.begin(),uy.end()),uy.end());
		for(int i = 0 ; i <= uy.size() ; i++) seg[i] = HappySequence();
		sort(ps.begin(),ps.end(),[&](array<int,3> a,array<int,3> b){
			if( a[0] != b[0] ) return a[0] < b[0];
			return make_pair(a[1],a[2]) > make_pair(b[1],b[2]);
		});
		int ans = 0;
		for( auto p : ps ){
			p[1] = lower_bound(uy.begin(),uy.end(),p[1]) - uy.begin() + 1;
			int mx = get(p[1]-1,p[2]) + 1;
			ans = max(ans,mx);
			ins(p[1],p[2],mx);
		}
		cout << ans << endl;
	}
}