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;

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;
}

struct Point {
	int x, y, z;
	Point(int x, int y, int z) : x(x), y(y), z(z) {}
	Point(){}
	bool operator < (const Point &a) const {
		return (x == a.x ? y == a.y ? z > a.z : y > a.y : x < a.x);
	}
};

struct Point2 {
	int y, z;
	Point2(int y, int z) : y(y), z(z) {}
	Point2(){}
	bool operator < (const Point2 &a) const {
		return (y < a.y);
	}
};

int main()
{
	int m, n, A, B;

	while (scanf("%d %d %d %d", &m, &n, &A, &B) && m + n){
		a = A; b = B;
		vector<Point> v(m + n);
        vector<set<Point2>> s(m + n);
		for (int i = 0; i < m; i++){
			int x, y, z;
			scanf("%d %d %d", &x, &y, &z);
			v[i] = Point(x, y, z);
		}
		for (int i = 0; i < n; i++){
			int x = r();
			int y = r();
			int z = r();
			v[i + m] = Point(x, y, z);
		}
		sort(v.begin(), v.end());

        int ans = 1;
        s[0].insert(Point2(v[0].y, v[0].z));
        for (int i = 1; i < m + n; i++){
            int lf = -1, rg = ans - 1;
            while (lf != rg){
                int mid = (lf + rg + 1) / 2;
                auto it = s[mid].lower_bound(Point2(v[i].y, v[i].z));
                if (it != s[mid].begin() && (--it)->z < v[i].z) lf = mid;
                else rg = mid - 1;
            }
            rg++;
            ans = max(ans, rg + 1);

            auto it = s[rg].lower_bound(Point2(v[i].y, v[i].z));
            while (it != s[rg].end() && it->z >= v[i].z) it = s[rg].erase(it);

            if (!s[rg].size()) s[rg].insert(Point2(v[i].y, v[i].z));
            else {
                auto it = s[rg].lower_bound(Point2(v[i].y, v[i].z));
                if (it == s[rg].begin() || (--it)->z != v[i].z) s[rg].insert(Point2(v[i].y, v[i].z));
            }
        }
        printf("%d\n", ans);
	}

	return (0);
}