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;
constexpr int C = ~(1 << 31);
constexpr 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;
}

struct point {
    int y, z;

    point(int y_, int z_) : y(y_), z(z_) {}
    bool operator<(point const& p) const {
        return y < p.y;
    }
};

int main() {
    ios_base::sync_with_stdio(false); cin.tie(0);

    int m, n;
    while(cin >> m >> n >> a >> b, n + m) {
        vector<tuple<int, int, int>> ps(n + m);
        for(int i = 0; i < m; ++i) {
            int x, y, z; cin >> x >> y >> z;
            ps[i] = make_tuple(x, -y, -z);
        }
        for(int i = m; i < n + m; ++i) {
            const int x = r(), y = r(), z = r();
            ps[i] = make_tuple(x, -y, -z);
        }
        sort(begin(ps), end(ps));
        vector<point> ps2;
        for(int i = 0; i < n + m; ++i) {
            ps2.emplace_back(-get<1>(ps[i]), -get<2>(ps[i]));
        }

        int ans = 0;
        vector<set<point>> dp(n + m + 1);
        dp[0].emplace(0, 0);
        for(auto const& p : ps2) {
            auto check = [&] (int i) {
                auto it = dp[i].lower_bound(p);
                if(it == dp[i].begin()) return false;
                it = prev(it);
                return p.y > it->y && p.z > it->z;
            };
            int lb = 0, ub = n + m;
            while(ub - lb > 1) {
                const int mid = (ub + lb) >> 1;
                (check(mid) ? lb : ub) = mid;
            }
            auto it = dp[ub].lower_bound(p);
            while(it != dp[ub].end() && it->z >= p.z) {
                it = dp[ub].erase(it);
            }
            if(it == dp[ub].begin() || prev(it)->z > p.z) {
                dp[ub].insert(p);
            }
            ans = max(ans, ub);
        }

        cout << ans << endl;
    }
}