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;
typedef pair<int,int> P;
#define MAX 300005
int m,n,A,B;
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;
}

int x[MAX],y[MAX],z[MAX];
set< P > t[MAX];


void insert(set<P> &s,P p){
  set<P> :: iterator it;
  while(1){
    it=s.upper_bound(p);
    if(it==s.end())break;

    if( p.first <= it->first &&
        p.second <= it->second ){
      s.erase(it);
    }else{
      break;
    }
  }
  s.insert(p);
}

bool check(set<P> &s,P p){
  set<P> :: iterator it;
  it=s.upper_bound(p);
  int cc=0;
  while(it!=s.begin()){
    cc++;
    if(cc>3)break;

    it--;
    if(it->first < p.first)
      if(it->second < p.second)
        return true;
  }
  return false;
}

int calc(P p){
  int left=0,right=m+n,mid;
  while(left<right){
    mid=(left+right)/2;
    if(check(t[mid],p))left=mid+1;
    else right=mid;
  }
  return left;
}


vector<P> G[1000005];

int solve(){
  int ans=1;

  for(int i=0;i<=m+n;i++){
    t[i].clear();
  }
  t[0].insert(P(-1,-1));
  set<int> st;

  for(int i=0;i<m+n;i++){
    st.insert( z[i] );
    G[ z[i] ].push_back( P(x[i],y[i]) );
  }

  for( int i : st ){
    vector<int> v( G[i].size() );
    for(int j=0;j<(int)G[i].size();j++){
      v[j]=calc(G[i][j]);
      ans=max(ans,v[j]);
    }
    for(int j=0;j<(int)G[i].size();j++){
      if(check( t[ v[j] ] , G[i][j])==false)
        insert(t[ v[j] ],G[i][j]);
    }
    G[i].clear();    
  }
  /*
  for(int i=0;i<=ans;i++){
    cout<<i<<" :"<<endl;
    for(P p:t[i]){
      cout<<p.first<<' '<<p.second<<endl;
    }
    cout<<endl;
    cout<<endl;
  }
  cout<<endl;
  */
  return ans;
}

int main(){
  while(1){
    scanf("%d %d %d %d",&m,&n,&A,&B);
    if(m==0&&n==0&&A==0&&B==0)break;

    a=A,b=B;
    
    for(int i=0;i<m;i++){
      scanf("%d %d %d",&x[i],&y[i],&z[i]);
    }

    for(int i=m;i<m+n;i++){
      x[i]=r();
      y[i]=r();
      z[i]=r();
    }
    printf("%d\n",solve());


  }
}