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

using namespace std;
struct Generator{
    int a, b, C, M;
    Generator(int A, int B){
        a = A, b = B, C = ~(1<<31), M = (1<<16)-1;
    }
    int generate() {
      a = 36969 * (a & M) + (a >> 16);
      b = 18000 * (b & M) + (b >> 16);
      return (C & ((a << 16) + b)) % 1000000;
    }
};

const int INF = 100000000;
typedef pair<int, int> Point;

void insert(int x, int y, set<Point>& s){
    const Point np(x, y);
    auto itr = s.lower_bound(np);
    if((--itr)->second <= y) return ;
    itr++;
    vector<Point> removes;
    while(itr->second >= y){
        removes.push_back(*itr);
        itr++;
    }
    for(Point p : removes){
        s.erase(p);
    }
    s.insert(np);
    /*
    for(auto p : s){
        printf("(%d, %d) ", p.first, p.second);
    }
    cout << endl;
    */
}

int main(){
    int M, N, A, B;
    const int MAX = 1000000;
    while(cin >> M >> N >> A >> B && (M + N) != 0){
        vector<vector<Point>> rect(MAX);
        for(int i = 0; i < M; i++){
            int x;
            Point p;
            cin >> x >> p.first >> p.second;
            rect[x].push_back(p);
        }
        Generator gen(A, B);
        for(int i = 0; i < N; i++){
            Point p;
            int x = gen.generate();
            p.first = gen.generate();
            p.second = gen.generate();
            rect[x].push_back(p);
        }

        vector<set<Point>> sets(N + M + 1);
        for(auto& s : sets){
            s.insert(Point(INF, -1));
            s.insert(Point(-1, INF));
        }

        int answer = 0;
        for(const vector<Point>& points : rect){
            vector<pair<int, Point>> querys;
            for(const Point& p : points){
                int x = p.first, y = p.second;
                int l = -1, r = answer + 1;
                while(r - l > 1){
                    int c = (l + r) / 2;
                    auto itr = sets[c].lower_bound(Point(x, -INF));
                    if((--itr)->second < y){
                        l = c;
                    }else{
                        r = c;
                    }
                }
                querys.emplace_back(r, p);
                answer = max(answer, r + 1);
            }
            for(const auto& query : querys){
                int k = query.first;
                Point p = query.second;
                //printf("%d (%d, %d)\n", k, p.first, p.second);
                insert(p.first, p.second, sets[k]);
            }
        }
        cout << answer << endl;
    }
    return 0;
}