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<iostream>
#include<algorithm>
#include<map>
#include<set>
#include<vector>
using namespace std;
#define INF (1 << 30)
typedef pair<int, int> P;



int C = ~(1<<31), M = (1<<16)-1;
int m, n, a, b, x, y, z;
int r() {
  a = 36969 * (a & M) + (a >> 16);
  b = 18000 * (b & M) + (b >> 16);
  return (C & ((a << 16) + b)) % 1000000;
}

template <typename T>
T next(T a){
	return ++a;
}

template <typename T>
T before(T a){
	return --a;
}

vector<P> points[1080000];
set<P> dp[324000];

int main(){
	while(cin >> m >> n >> a >> b, m | n | a | b){
		for(int i = 0;i < m;i++){
			cin >> x >> y >> z;
			points[x].push_back(P(-y, z));
		}
		for(int i = 0;i < n;i++){
			x = r();
			y = r();
			z = r();
			points[x].push_back(P(-y, z));
		}
		n += m;
		for(int i = 0;i < 324000;i++){
			while(!dp[i].empty())dp[i].erase(dp[i].begin());
		}
		dp[0].insert(P(-INF, -INF));
        int res = 0;
		for(int i = 0;i < 1080000;i++){
			sort(points[i].begin(), points[i].end());
			for(int j = 0;j < points[i].size();j++){
				y = -points[i][j].first;
				z = points[i][j].second;
				int bottom = 0, top = 324000;
				while(top - bottom > 1){
					int mid = (top + bottom) / 2;
					if(dp[mid].empty()){
						top = mid;
						continue;
					}
					set<P>::iterator it = dp[mid].lower_bound(P(y, -INF));					
					if(it == dp[mid].begin()){
						top = mid;
						continue;
					}
					it--;
					if(it->second < z)bottom = mid;
					else top = mid;
				}
				bottom++;
				res = max(res, bottom);
				dp[bottom].insert(P(y, z));
				set<P>::iterator it = dp[bottom].find(P(y, z));

				if(it != dp[bottom].begin() && before(it)->second <= z){
					dp[bottom].erase(it);
					continue;
				}
				it++;
				while(it != dp[bottom].end() && it->second >= z){
					dp[bottom].erase(it++);
				}			
			}
			points[i].clear();
		}
		cout << res << endl;
	}	
	return 0;
}