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;

#define lp(i, a, b) for (int i = (int)(a); i < (int)(b); i++)
#define rep(i, n) lp(i, 0, n)
#define fi first
#define se second

const int INF = 1 << 28;

typedef pair<int, int> pr;
typedef tuple<int, int, int> tp;

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;
}

vector<tp> ps;

int solve()
{
	rep(i, ps.size()) get<1>(ps[i]) *= -1;
	sort(begin(ps), end(ps));
	rep(i, ps.size()) get<1>(ps[i]) *= -1;
	
	vector<set<pr>> sts(ps.size() + 10);
	for (auto &st : sts){
		st.insert(pr(-INF, INF));
		st.insert(pr(INF, -INF));
	}
	
	int res = 0;
	rep(i, ps.size()){
		int x, y, z;
		tie(x, y, z) = ps[i];
		int l = 0, r = ps.size();
		while (l + 1 < r){
			int m = (l + r) / 2;
			auto it = --sts[m].lower_bound(pr(y, -INF));
			if (it->se < z) l = m;
			else r = m;
		}
		if (z <= (--sts[r].lower_bound(pr(y, z)))->se){
			while (true){
				auto it = sts[r].lower_bound(pr(y, z));
				if (it->se < z) break;
				sts[r].erase(it);
			}
			sts[r].insert(pr(y, z));
		}
		res = max(res, r);
	}

	return res;
}

int main()
{
	int m, n;
	while (cin >> m >> n >> a_ >> b_, m + n){
		ps.clear();
		rep(i, m){
			int x, y, z;
			cin >> x >> y >> z;
			ps.emplace_back(x, y, z);
		}
		rep(i, n){
			int x = r_();
			int y = r_();
			int z = r_();
			ps.emplace_back(x, y, z);
		}
		cout << solve() << endl;
	}
}