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(){}
  Point(int x, int y, int z):x(x), y(y), z(z){}
  bool operator == (const Point& rhs) const {
    return x==rhs.x && y==rhs.y && z==rhs.z;
  }
  bool operator < (const Point& rhs) const {
    if(x != rhs.x) return x < rhs.x;
    if(y != rhs.y) return y > rhs.y;
    return z > rhs.z;
  }
};

struct Point2D {
  int y, z;
  Point2D(){}
  Point2D(int y, int z):y(y), z(z){}
  bool operator == (const Point2D& rhs) const {
    return y==rhs.y;
  }
  bool operator < (const Point2D& rhs) const {
    return y < rhs.y;
  }
};


int main() {
  int m, n, A, B;
  while(cin >> m >> n >> A >> B, m+n) {
    a = A, b = B;

    int N = n+m;
    vector<Point> pts(N);
    for(int i = 0; i < m; ++i) {
      cin >> pts[i].x >> pts[i].y >> pts[i].z;
    }
    for(int i = m; i < N; ++i) {
      pts[i].x = r();
      pts[i].y = r();
      pts[i].z = r();
    }
    sort(pts.begin(), pts.end());

    vector<set<Point2D> > dp(N+1);
    int ans = 1;
    for(int i = 0; i < N; ++i) {
      Point2D p = Point2D(pts[i].y, pts[i].z);
      int lb = 0, ub = ans+1;
      while(lb+1 < ub) {
	int mb = (lb+ub)/2;
	auto it = dp[mb].lower_bound(p);
	if(it != dp[mb].begin() && (--it)->z < p.z) lb = mb;
	else ub = mb;
      }
      ans = max(ans, ub);
      if(!dp[ub].empty()) {
	auto it = dp[ub].lower_bound(p);
	while(it != dp[ub].end() && it->z >= p.z) {
	  it = dp[ub].erase(it);
	}
      }
      if(dp[ub].empty()) {
	dp[ub].insert(p);
      } else {
	auto it = dp[ub].lower_bound(p);
	if(it == dp[ub].begin() || (--it != dp[ub].end() && (it->y != p.y || it->z > p.z))) {
	  dp[ub].insert(p);
	}
      }
    }

    cout << ans << endl;
  }

  return 0;
}