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 <vector>
#include <algorithm>
using namespace std;
int N, M, A, B, C = ~(1 << 31), D = 65535, lim = 1000001, x[300009], y[300009], z[300009], p[300009];
vector<int> cp[1000009], bit[1000009];
int func() {
	A = 36969 * (A & D) + (A >> 16);
	B = 18000 * (B & D) + (B >> 16);
	return (C & ((A << 16) + B)) % 1000000;
}
void assign(int px, int py, int val) {
	for (int i = px + 1; i <= lim; i += i & (-i)) {
		int ptr = lower_bound(cp[i].begin(), cp[i].end(), py) - cp[i].begin();
		for (int j = ptr + 1; j <= cp[i].size(); j += j & (-j)) {
			bit[i][j - 1] = max(bit[i][j - 1], val);
		}
	}
}
int query(int px, int py) {
	int ret = 0;
	for (int i = px; i >= 1; i -= i & (-i)) {
		int ptr = lower_bound(cp[i].begin(), cp[i].end(), py) - cp[i].begin();
		for (int j = ptr; j >= 1; j -= j & (-j)) {
			ret = max(ret, bit[i][j - 1]);
		}
	}
	return ret;
}
int main() {
	while (scanf("%d%d%d%d", &N, &M, &A, &B), N + M) {
		for (int i = 0; i < N; ++i) scanf("%d%d%d", &x[i], &y[i], &z[i]);
		for (int i = 0; i < M; ++i) x[N] = func(), y[N] = func(), z[N] = func(), ++N;
		for (int i = 0; i < N; ++i) p[i] = i;
		sort(p, p + N, [](int i, int j) { return z[i] != z[j] ? z[i] < z[j] : y[i] > y[j]; });
		for (int i = 0; i <= lim; i++) cp[i].clear();
		for (int i = 0; i < N; i++) {
			for (int j = x[p[i]] + 1; j <= lim; j += j & (-j)) {
				cp[j].push_back(y[p[i]]);
			}
		}
		for (int i = 0; i <= lim; i++) {
			sort(cp[i].begin(), cp[i].end());
			cp[i].erase(unique(cp[i].begin(), cp[i].end()), cp[i].end());
			bit[i] = vector<int>(cp[i].size(), 0);
		}
		int ret = 0;
		for (int i = 0; i < N; i++) {
			int res = query(x[p[i]], y[p[i]]);
			assign(x[p[i]], y[p[i]], res + 1);
			ret = max(ret, res + 1);
		}
		printf("%d\n", ret);
	}
	return 0;
}