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 <map>
#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;
#define FOR(i,n) for(int i = 0; i < (n); i++)
#define sz(c) ((int)(c).size())

int m, n, a, b;

struct P {
	int x, y, z;
	P() {}
	P(int x, int y, int z) : x(x), y(y), z(z) {}
	bool operator< (const P& l) const {
		if (x != l.x) return x < l.x;
		if (y != l.y) return y > l.y;
		return z > l.z;
	}
};

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

bool C(map<int, P>& polyline, const P& p) {
	map<int, P>::iterator it = polyline.lower_bound(p.y);
	if (it == polyline.begin()) return true;
	it--;
	return it->second.z >= p.z;
}

void insert_polyline(vector<map<int, P> >& polyline, const P& p) {
	int l = -1, r = sz(polyline);
	while (r - l != 1) {
		int md = (l + r) / 2;
		bool ok = C(polyline[md], p);
		if (ok) r = md;
		else l = md;
	}
	if (sz(polyline) == r) polyline.emplace_back();
	if (polyline[r].find(p.y) != polyline[r].end() && polyline[r][p.y].z <= p.z) return;
	polyline[r][p.y] = p;
	map<int, P>::iterator it = polyline[r].find(p.y);
	++it;
	while (it != polyline[r].end()) {
		if (it->second.z < p.z) break;
		polyline[r].erase(it++);
	}
}

int main() {
	int cnt = 0;
	while (cin >> m >> n >> a >> b, m || n || a || b) {
		cnt++;
		vector<P> v;
		FOR(i, m) {
			int x, y, z; cin >> x >> y >> z;
			v.emplace_back(x, y, z);
		}
		FOR(i, n) {
			int x = r();
			int y = r();
			int z = r();
			v.emplace_back(x, y, z);
		}
		sort(v.begin(), v.end());

		vector<map<int, P> > polyline;
		FOR(i, sz(v)) {
			insert_polyline(polyline, v[i]);
		}

		cout << sz(polyline) << "\n";
	}

	return 0;
}