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 REP(i,n) for(int i=0; i<(int)(n); i++)

using ll = long long;

int a,b;
// const int C = ~(1<<31);
const int C = (1<<30) + ((1<<30) - 1);
const int M = (1<<16)-1;
int r() {
	a = 36969 * (a & M) + (a >> 16);
	b = 18000 * (b & M) + (b >> 16);
	return (C & ((a << 16) + b)) % 1000000;
}

using pii = pair<int,int>;

pair<int,pii> points[325252];

int main(){
	while(true){
		int n,m;
		scanf("%d%d%d%d",&n,&m,&a,&b);
		if(n+m+a+b==0)break;
		REP(i,n){
			int x,y,z;
			scanf("%d%d%d",&x,&y,&z);
			x++; y++; z++;
			points[i] = make_pair(x,pii(y,z));
		}
		REP(i,m){
			int x = r();
			int y = r();
			int z = r();
			x++; y++; z++;
			points[i+n] = make_pair(x,pii(y,z));
		}
		n += m;
		sort(points, points+n, [&](pair<int,pii> a, pair<int,pii> b){
			if(a.first != b.first)return a.first < b.first;
			return a.second > b.second;
		});
		const int N = 1000025;
		std::unordered_map<int,std::map<int,int>> bit;
		auto add = [&](int y, int z, int v){
			for(; y<N; y+=y&-y){
				// add to y
				if(bit.count(y)==0){
					bit[y] = std::map<int,int>();
					bit[y][-1] = -252521;
				}
				std::map<int,int> &mp = bit[y];
				auto it = mp.upper_bound(z); it--;
				if(it->second >= v)continue;
				mp[z] = v;
				it = mp.upper_bound(z);
				while(it != mp.end() && it->second <= v){
					mp.erase(it++);
				}
			}
		};
		auto query = [&](int y, int z){
			int ans = 0;
			for(; y>0; y-=y&-y){
				// get from y
				if(bit.count(y)==0)continue;
				std::map<int,int> &mp = bit[y];
				auto it = mp.upper_bound(z); it--;
				ans = max(ans, it->second);
			}
			return ans;
		};
		int ans = 0;
		REP(i,n){
			int x = points[i].first;
			int y = points[i].second.first;
			int z = points[i].second.second;
			int v = query(y-1, z-1) + 1;
			ans = max(ans, v);
			add(y,z,v);
		}
		printf("%d\n",ans);
	}
	return 0;
}