#include <iostream>
#include <algorithm>
#include <queue>
#define ll long long
#define pii pair<int, int>
#define fst first
#define snd second
using namespace std;
int B, N, D, M;
namespace solve1D // O(N log N)
{
int A[100001];
inline ll solve()
{
ll ret = 0;
for (int i = 0; i < N; i++) cin >> A[i];
sort(A, A + N);
int j = 0;
for (int i = 1; i < N; i++)
{
while (j < i && A[j] + D < A[i]) {j++;}
ret += i - j;
}
return ret;
}
}
namespace solve2D // O(N log M)
{
pii A[100001]; int BIT[150001];
inline void upd(int x, int v)
{
for (; x <= 2 * M; x += x & (-x)) {BIT[x] += v;}
}
inline int qry(int x)
{
int ret = 0;
x = min(x, 2 * M); x = max(x, 0);
for (; x; x -= x & (-x)) {ret += BIT[x];}
return ret;
}
inline ll solve()
{
ll ret = 0;
for (int i = 0; i < N; i++)
{
int x, y; cin >> x >> y;
A[i] = {x - y, x + y};
}
sort(A, A + N);
int j = 0;
for (int i = 0; i < N; i++)
{
while (j < i && A[j].fst + D < A[i].fst)
upd(A[j++].snd, -1);
ret += qry(A[i].snd + D) - qry(A[i].snd - D - 1);
upd(A[i].snd, 1);
}
return ret;
}
}
namespace solve3D // O(M^3 + NM)
{
int A[100001][3], pref[80][160][160];
inline void process(int l)
{
for (int i = 0; i <= 2 * M; i++)
{
for (int j = 0; j <= 2 * M; j++)
{
if (i) pref[l][i][j] += pref[l][i - 1][j];
if (j) pref[l][i][j] += pref[l][i][j - 1];
if (i && j) pref[l][i][j] -= pref[l][i - 1][j - 1];
}
}
}
inline int qry(int l, int x1, int y1, int x2, int y2)
{
x1 = max(x1 - 1, 0); y1 = max(y1 - 1, 0);
x2 = min(x2, 2 * M); y2 = min(y2, 2 * M);
return pref[l][x2][y2] - pref[l][x1][y2] - pref[l][x2][y1] + pref[l][x1][y1];
}
inline ll solve()
{
ll ret = 0;
for (int i = 0; i < N; i++)
{
int x, y; cin >> A[i][0] >> x >> y;
A[i][1] = x - y + M; A[i][2] = x + y;
pref[A[i][0]][A[i][1]][A[i][2]]++;
}
for (int i = 1; i <= M; i++) process(i);
for (int i = 0; i < N; i++)
for (int j = 1; j <= M; j++)
{
int d = D - abs(j - A[i][0]);
if (d >= 0) ret += qry(j, A[i][1] - d, A[i][2] - d, A[i][1] + d, A[i][2] + d);
}
return (ret - N) / 2;
}
}
int main()
{
ios :: sync_with_stdio(0); cin.tie(0); cout.tie(0);
cin >> B >> N >> D >> M;
if (B == 1) cout << solve1D :: solve() << "\n";
else if (B == 2) cout << solve2D :: solve() << "\n";
else cout << solve3D :: solve() << "\n";
return 0;
}
