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>
#include <map>
#include <array>
#include <utility>
#include <climits>
using namespace std;

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

int main(){
	int m, n, a, b, x, y, z;
	while(scanf("%d%d%d%d", &m, &n, &a, &b), m | n){
		vector<array<int,3>> v(n + m);
		for(int i = 0; i < m; ++i){
			scanf("%d%d%d", &x, &y, &z);
			v[i][0] = x;
			v[i][1] = -y;
			v[i][2] = -z;
		}
		for(int i = m; i < n + m; ++i){
			v[i][0] = rnd(a, b);
			v[i][1] = -rnd(a, b);
			v[i][2] = -rnd(a, b);
		}
		sort(v.begin(), v.end());
		int ans = 0;
		vector<map<int,int>> mps(m + n + 3);
		mps[0][100] = 100;
		for(size_t i = 1; i < mps.size(); ++i){
			mps[i][INT_MIN] = INT_MAX;
		}
		for(int i = 0; i < m + n; ++i){
			y = v[i][1];
			z = v[i][2];
			int left = 1, right = ans + 1;
			while(left < right){
				int mid = (left + right + 1) >> 1;
				const auto &mpm = mps[mid - 1];
				auto it = mpm.upper_bound(y);
				if(it != mpm.end() && it->second > z){
					left = mid;
				}
				else{
					right = mid - 1;
				}
			}
			
			auto &mpl = mps[left];
			for(auto it = mpl.upper_bound(y); ; ){
				--it;
				if(it->second > z){
					break;
				}
				it = mpl.erase(it);
			}
			
			int &t = mpl.insert(make_pair(y, z)).first->second;
			t = max(t, z);
			ans = max(ans, left);
		}
		printf("%d\n", ans);
	}
}