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;

const int MAXN = 300000 + 10;

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 || (x == rhs.x && y < rhs.y) || (x == rhs.x && y == rhs.y && z < rhs.z);
    }
} P[MAXN];

map<int, int> best[MAXN];
int dp[MAXN], ret;

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

inline bool check(const map<int, int> &mp, int x, int y) {
    auto it = mp.lower_bound(x);
    return it != mp.begin() && (-- it)->second < y;
}

int query(int x, int y) {
    int left = 0, right = ret;
    while (left < right) {
        int mid = (left + right + 1) >> 1;
        if (check(best[mid], x, y)) left = mid;
        else right = mid - 1;
    }
    return left;
}

void update(map<int, int> &mp, int x, int y) {
    auto it = mp.upper_bound(x);
    if (it != mp.begin() && (-- it)->second <= y) return;
    it = mp.lower_bound(x);
    while (it != mp.end() && it->second >= y) mp.erase(it ++);
    mp[x] = y;
}

int main() {
    int N, M, A, B;
    while (scanf("%d%d%d%d", &M, &N, &A, &B) == 4 && (N + M)) {
        for (int i = 0; i < M; ++ i) {
            scanf("%d%d%d", &P[i].x, &P[i].y, &P[i].z);
        }
        N += M;
        for (int i = M; i < N; ++ i) {
            P[i].x = r(A, B);
            P[i].y = r(A, B);
            P[i].z = r(A, B);
        }
        for (int i = 0; i <= N; ++ i) best[i].clear();
        sort(P, P + N);
        ret = 0;
        for (int i = 0, j; i < N;) {
            for (j = i; j < N && P[i].x == P[j].x; ++ j) dp[j] = query(P[j].y, P[j].z) + 1;
            for (; i < j; ++ i) {
                update(best[dp[i]], P[i].y, P[i].z);
                ret = max(ret, dp[i]);
            }
        }
        printf("%d\n", ret);
    }
    return 0;
}