Problem: In the last mission, MDCS has successfully shipped N AI robots to Mars. Before they start exploring, system initialization is required so they are arranged in a line. Every robot can be described with three numbers: position (x_i), radius of sight (r_i) and IQ (q_i).

Since they are intelligent robots, some of them will talk if they see each other. Radius of sight is inclusive, so robot can see other all robots in range [x_i - r_i, x_i + r_i]. But they don't walk to talk with anybody, but only with robots who have similar IQ. By similar IQ we mean that their absolute difference isn't more than K. 

Help us and calculate how many pairs of robots are going to talk with each other, so we can timely update their software and avoid any potential quarrel.

Input

The first line contains two integers, numbers N (1 ≤ N ≤ 10^5)  and K (0 ≤ K ≤ 20).

Next N lines contain three numbers each x_i, r_i, q_i (0 ≤ x_i,r_i,q_i ≤ 10^9) — position, radius of sight and IQ of every robot respectively.

Output

Output contains only one number — solution to the problem.

Example

Input

3 2
3 6 1
7 3 10
10 5 8


Output

1

Note

The first robot can see the second, but not vice versa. The first robot can't even see the third. The second and the third robot can see each other and their IQs don't differ more than 2 so only one conversation will happen.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma warning(disable : 4996)
using namespace std;
namespace Xrocks {}
using namespace Xrocks;
namespace Xrocks {
class in {
} user_input;
class out {
} output;
in& operator>>(in& X, int& Y) {
  scanf("%d", &Y);
  return X;
}
in& operator>>(in& X, char* Y) {
  scanf("%s", Y);
  return X;
}
in& operator>>(in& X, float& Y) {
  scanf("%f", &Y);
  return X;
}
in& operator>>(in& X, double& Y) {
  scanf("%lf", &Y);
  return X;
}
in& operator>>(in& X, char& C) {
  scanf("%c", &C);
  return X;
}
in& operator>>(in& X, string& Y) {
  cin >> Y;
  return X;
}
in& operator>>(in& X, long long& Y) {
  scanf("%lld", &Y);
  return X;
}
template <typename T>
in& operator>>(in& X, vector<T>& Y) {
  for (auto& x : Y) user_input >> x;
  return X;
}
template <typename T>
out& operator<<(out& X, const T& Y) {
  cout << Y;
  return X;
}
template <typename T>
out& operator<<(out& X, vector<T>& Y) {
  for (auto& x : Y) output << x << " ";
  return X;
}
out& operator<<(out& X, const int& Y) {
  printf("%d", Y);
  return X;
}
out& operator<<(out& X, const char& C) {
  printf("%c", C);
  return X;
}
out& operator<<(out& X, const string& Y) {
  printf("%s", Y.c_str());
  return X;
}
out& operator<<(out& X, const long long& Y) {
  printf("%lld", Y);
  return X;
}
out& operator<<(out& X, const float& Y) {
  printf("%f", Y);
  return X;
}
out& operator<<(out& X, const double& Y) {
  printf("%lf", Y);
  return X;
}
out& operator<<(out& X, const char Y[]) {
  printf("%s", Y);
  return X;
}
template <typename T>
T max(T A) {
  return A;
}
template <typename T, typename... args>
T max(T A, T B, args... S) {
  return max(A > B ? A : B, S...);
}
template <typename T>
T min(T A) {
  return A;
}
template <typename T, typename... args>
T min(T A, T B, args... S) {
  return min(A < B ? A : B, S...);
}
template <typename T>
void vectorize(int y, vector<T>& A) {
  A.resize(y);
}
template <typename T, typename... args>
void vectorize(int y, vector<T>& A, args&&... S) {
  A.resize(y);
  vectorize(y, S...);
}
long long fast(long long a, long long b, long long pr) {
  if (b == 0) return 1 % pr;
  long long ans = 1 % pr;
  while (b) {
    if (b & 1) ans = (ans * a) % pr;
    b >>= 1;
    a = (a * a) % pr;
  }
  return ans;
}
int readInt() {
  int n = 0;
  int ch = getchar_unlocked();
  int sign = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') sign = -1;
    ch = getchar_unlocked();
  }
  while (ch >= '0' && ch <= '9')
    n = (n << 3) + (n << 1) + ch - '0', ch = getchar_unlocked();
  n = n * sign;
  return n;
}
long long readLong() {
  long long n = 0;
  int ch = getchar_unlocked();
  int sign = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') sign = -1;
    ch = getchar_unlocked();
  }
  while (ch >= '0' && ch <= '9')
    n = (n << 3) + (n << 1) + ch - '0', ch = getchar_unlocked();
  n = n * sign;
  return n;
}
long long readBin() {
  long long n = 0;
  int ch = getchar_unlocked();
  int sign = 1;
  while (ch < '0' || ch > '1') {
    if (ch == '-') sign = -1;
    ch = getchar_unlocked();
  }
  while (ch >= '0' && ch <= '1')
    n = (n << 1) + (ch - '0'), ch = getchar_unlocked();
  return n;
}
long long inv_(long long val,
               long long pr = static_cast<long long>(1000000007)) {
  return fast(val, pr - 2, pr);
}
}  // namespace Xrocks
int capK, capK2;
class seg_tree {
  vector<int> tree;
  vector<map<int, int>> tree2;
  int n;
  inline int get_size(int R) {
    R--;
    int Z = 1;
    while (R) {
      Z <<= 1;
      R >>= 1;
    }
    return Z * 2;
  }

 public:
  void construct(int s) {
    tree2.resize(get_size(s));
    n = s - 1;
  }
  int find_max(int l, int r, int K) { return find_max2(l, r, K, 0, n, 1); }
  int find_max2(int l, int r, int K, int L, int R, int curr) {
    if (r < L || l > R || tree2[curr].lower_bound(K) == tree2[curr].end())
      return 0;
    if (L >= l && R <= r) {
      int ans = 0;
      auto i = tree2[curr].lower_bound(K);
      while (i != tree2[curr].end() && i->first <= K + capK2) {
        ans += i->second;
        ++i;
      }
      return ans;
    }
    int c = curr;
    return (find_max2(l, r, K, L, (L + R) / 2, 2 * curr) +
            find_max2(l, r, K, (L + R) / 2 + 1, R, 2 * curr + 1));
  }
  void update(int pos, int val) { update(pos, val, 1, 0, n); }
  void update(int pos, int val, int c, int L, int R) {
    int curr = c;
    if (pos < L || pos > R) return;
    if (L == R) {
      tree2[c][val]++;
      return;
    }
    update(pos, val, curr << 1, L, (L + R) >> 1);
    update(pos, val, (curr << 1) + 1, ((L + R) >> 1) + 1, R);
    tree2[c][val]++;
    return;
  }
  void update2(int pos, int val) { update2(pos, val, 1, 0, n); }
  void update2(int pos, int val, int c, int L, int R) {
    int curr = c;
    if (pos < L || pos > R) return;
    if (L == R) {
      tree2[c][val]--;
      return;
    }
    update2(pos, val, curr << 1, L, (L + R) >> 1);
    update2(pos, val, (curr << 1) + 1, ((L + R) >> 1) + 1, R);
    tree2[c][val]--;
    return;
  }
};
class solve {
  long long n, k;
  vector<int> v, r, q, invw;
  seg_tree St;

 public:
  solve() {
    user_input >> n >> capK;
    capK2 = 2 * capK;
    v.resize(n);
    r.resize(n);
    q.resize(n);
    invw.resize(n);
    vector<int> idx(n);
    for (int i = 0; i < n; i++) {
      idx[i] = i;
      user_input >> v[i] >> r[i] >> q[i];
    }
    sort(idx.begin(), idx.end(),
         [&](const int& a, const int& b) { return v[a] < v[b]; });
    vector<int> v2(1);
    invw[idx[0]] = 0;
    v2[0] = v[idx[0]];
    int prev = v2[0];
    for (int i = 1; i < n; i++) {
      if (v[idx[i - 1]] == v[idx[i]]) {
        invw[idx[i]] = invw[idx[i - 1]];
      } else {
        invw[idx[i]] = invw[idx[i - 1]] + 1;
      }
      if (prev != v[idx[i]]) v2.push_back(v[idx[i]]);
      prev = v[idx[i]];
    }
    St.construct(v2.size());
    set<pair<int, int>> S;
    long long a = 0;
    for (int i = 0; i < n; i++) {
      int c = idx[i];
      while (!S.empty() && S.begin()->first < v[c]) {
        St.update2(invw[S.begin()->second], q[S.begin()->second]);
        S.erase(S.begin());
      }
      int l2, r2;
      l2 = lower_bound(v2.begin(), v2.end(), v[c] - r[c]) - v2.begin();
      r2 = upper_bound(v2.begin(), v2.end(), v[c] + r[c]) - v2.begin();
      --r2;
      a += St.find_max(l2, r2, q[c] - capK);
      St.update(invw[c], q[c]);
      S.insert({v[c] + r[c], c});
    }
    output << a;
  }
  int gcd(int a, int b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
};
int32_t main() {
  int t = 1, i = 1;
  if (0 || 0) scanf("%d", &t);
  while (t--) {
    if (0) printf("Case #%d: ", i++);
    new solve;
  }
  output << "\n";
  return 0;
}