Let us compare two triples a = (xa, ya, za) and b = (xb, yb, zb) by a partial order ∠ defined as follows.
Your mission is to find, in the given set of triples, the longest ascending series a1 ∠ a2 ∠ ... ∠ ak.
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.
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.
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
3 5 1 3