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 JAVA):

Solution:
import java.util.Arrays;
import java.util.Comparator;
import java.util.Scanner;
import java.util.TreeMap;
import java.util.TreeSet;

class Main {
	public static void main(String[] args) {
		new Main().run();
	}

	boolean check(TreeSet<int[]> C, int[] S) {
		int[] Sl = C.lower(S);
		if (Sl != null && Sl[1] < S[1]) {
			return true;
		} else {
			return false;
		}
	}

	@SuppressWarnings("unchecked")
	void solve(int N, int[][] S) {
		TreeSet<int[]>[] C = new TreeSet[N];
		for (int i = 0; i < N; ++i) {
			C[i] = new TreeSet<>(new Comparator<int[]>() {
				@Override
				public int compare(int[] o1, int[] o2) {
					if (o1[0] != o2[0]) {
						return Integer.compare(o1[0], o2[0]);
					} else {
						return -Integer.compare(o1[1], o2[1]);
					}
				}
			});
		}
		C[0].add(S[0]);
		int maxLen = 1;
		for (int i = 1; i < N; ++i) {
			int left = -1;
			int right = maxLen;
			while (right - left > 1) {
				int middle = (right + left) / 2;
				if (check(C[middle], S[i])) {
					left = middle;
				} else {
					right = middle;
				}
			}

			if (left + 1 == maxLen) {
				C[left + 1].add(S[i]);
				++maxLen;
				continue;
			} else {
				int[] tmp = C[left + 1].floor(S[i]);
				if (tmp != null && tmp[1] == S[i][1])
					continue;
				tmp = C[left + 1].higher(S[i]);
				while (tmp != null && S[i][1] <= tmp[1]) {
					C[left + 1].remove(tmp);
					tmp = C[left + 1].higher(S[i]);
				}
				C[left + 1].add(S[i]);
			}
		}
		System.out.println(maxLen);
	}

	int nextA(int a, int M) {
		return 36969 * (a & M) + (a >> 16);
	}

	int nextB(int b, int M) {
		return 18000 * (b & M) + (b >> 16);
	}

	int nextVal(int a, int b, int C) {
		return (C & ((a << 16) + b)) % 1000000;
	}

	void run() {
		Scanner sc = new Scanner(System.in);
		while (true) {
			int m = sc.nextInt();
			int n = sc.nextInt();
			int A = sc.nextInt();
			int B = sc.nextInt();
			if (m == 0 && n == 0 && A == 0 && B == 0)
				break;
			int[][] S = new int[m + n][3];
			for (int i = 0; i < m; ++i) {
				S[i][0] = sc.nextInt();
				S[i][1] = sc.nextInt();
				S[i][2] = sc.nextInt();
			}
			int a = A, b = B, C = ~(1 << 31), M = (1 << 16) - 1;
			for (int i = 0; i < n; ++i) {
				for (int j = 0; j < 3; ++j) {
					a = nextA(a, M);
					b = nextB(b, M);
					S[i + m][j] = nextVal(a, b, C);
				}
			}
			Arrays.sort(S, new Comparator<int[]>() {
				@Override
				public int compare(int[] o1, int[] o2) {
					if (o1[0] != o2[0])
						return Integer.compare(o1[0], o2[0]);
					else if (o1[1] != o2[1]) {
						return -Integer.compare(o1[1], o2[1]);
					} else {
						return -Integer.compare(o1[2], o2[2]);
					}
				}
			});

			int[][] tmp = new int[m + n][2];
			for (int i = 0; i < m + n; ++i) {
				tmp[i][0] = S[i][1];
				tmp[i][1] = S[i][2];
			}
			solve(m + n, tmp);
		}
	}

	void tr(Object... objects) {
		System.out.println(Arrays.deepToString(objects));
	}
}