#include <bits/stdc++.h>
using namespace std;
struct edge {
  long long to, cost;
  edge() {}
  edge(long long a, long long b) { to = a, cost = b; }
};
const long long dx[] = {1, 0, -1, 0}, dy[] = {0, -1, 0, 1};
const long long mod = 1000000007;
struct mint {
  long long x = 0;
  mint(long long y = 0) {
    x = y;
    if (x < 0 || x >= mod) x = (x % mod + mod) % mod;
  }
  mint(const mint& ope) { x = ope.x; }
  mint operator-() { return mint(-x); }
  mint operator+(const mint& ope) { return mint(x) += ope; }
  mint operator-(const mint& ope) { return mint(x) -= ope; }
  mint operator*(const mint& ope) { return mint(x) *= ope; }
  mint operator/(const mint& ope) { return mint(x) /= ope; }
  mint& operator+=(const mint& ope) {
    x += ope.x;
    if (x >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint& ope) {
    x += mod - ope.x;
    if (x >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint& ope) {
    x *= ope.x, x %= mod;
    return *this;
  }
  mint& operator/=(const mint& ope) {
    long long n = mod - 2;
    mint mul = ope;
    while (n) {
      if (n & 1) *this *= mul;
      mul *= mul;
      n >>= 1;
    }
    return *this;
  }
  mint inverse() { return mint(1) / *this; }
  bool operator==(const mint& ope) { return x == ope.x; }
  bool operator!=(const mint& ope) { return x != ope.x; }
};
mint modpow(mint a, long long n) {
  if (n == 0) return mint(1);
  if (n % 2)
    return a * modpow(a, n - 1);
  else
    return modpow(a * a, n / 2);
}
istream& operator>>(istream& is, mint& ope) {
  long long t;
  is >> t, ope.x = t;
  return is;
}
ostream& operator<<(ostream& os, mint& ope) { return os << ope.x; }
ostream& operator<<(ostream& os, const mint& ope) { return os << ope.x; }
bool exceed(long long x, long long y, long long m) { return x >= m / y + 1; }
void mark() { cout << "*" << endl; }
void yes() { cout << "YES" << endl; }
void no() { cout << "NO" << endl; }
long long sgn(long long x) {
  if (x > 0) return 1;
  if (x < 0) return -1;
  return 0;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long digitnum(long long x, long long b = 10) {
  long long ret = 0;
  for (; x; x /= b) ret++;
  return ret;
}
long long digitsum(long long x, long long b = 10) {
  long long ret = 0;
  for (; x; x /= b) ret += x % b;
  return ret;
}
string lltos(long long x) {
  string ret;
  for (; x; x /= 10) ret += x % 10 + '0';
  reverse(ret.begin(), ret.end());
  return ret;
}
long long stoll(string& s) {
  long long ret = 0;
  for (auto c : s) ret *= 10, ret += c - '0';
  return ret;
}
template <typename T>
void uniq(T& vec) {
  sort(vec.begin(), vec.end());
  vec.erase(unique(vec.begin(), vec.end()), vec.end());
}
template <typename T>
ostream& operator<<(ostream& os, vector<T>& vec) {
  for (int i = 0; i < vec.size(); i++)
    os << vec[i] << (i + 1 == vec.size() ? "" : " ");
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, deque<T>& deq) {
  for (int i = 0; i < deq.size(); i++)
    os << deq[i] << (i + 1 == deq.size() ? "" : " ");
  return os;
}
template <typename T, typename U>
ostream& operator<<(ostream& os, pair<T, U>& pair_var) {
  os << "(" << pair_var.first << ", " << pair_var.second << ")";
  return os;
}
template <typename T, typename U>
ostream& operator<<(ostream& os, const pair<T, U>& pair_var) {
  os << "(" << pair_var.first << ", " << pair_var.second << ")";
  return os;
}
template <typename T, typename U>
ostream& operator<<(ostream& os, map<T, U>& map_var) {
  for (typename map<T, U>::iterator itr = map_var.begin(); itr != map_var.end();
       itr++) {
    os << "(" << itr->first << ", " << itr->second << ")";
    itr++;
    if (itr != map_var.end()) os << ",";
    itr--;
  }
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, set<T>& set_var) {
  for (typename set<T>::iterator itr = set_var.begin(); itr != set_var.end();
       itr++) {
    os << *itr;
    ++itr;
    if (itr != set_var.end()) os << " ";
    itr--;
  }
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, multiset<T>& set_var) {
  for (typename multiset<T>::iterator itr = set_var.begin();
       itr != set_var.end(); itr++) {
    os << *itr;
    ++itr;
    if (itr != set_var.end()) os << " ";
    itr--;
  }
  return os;
}
template <typename T>
void outa(T a[], long long s, long long t) {
  for (long long i = s; i <= t; i++) {
    cout << a[i];
    if (i < t) cout << " ";
  }
  cout << endl;
}
void dump_func() { cout << endl; }
template <class Head, class... Tail>
void dump_func(Head&& head, Tail&&... tail) {
  cout << head;
  if (sizeof...(Tail) > 0) cout << " ";
  dump_func(std::move(tail)...);
}
struct ArraybySet {
  set<pair<long long, long long> > S;
  ArraybySet() { init(); }
  void init() {
    S.clear();
    S.insert(pair<long long, long long>(-1e18, 0));
  }
  set<pair<long long, long long> >::iterator get(long long p) {
    auto it = S.upper_bound(pair<long long, long long>(p, 1e18));
    it--;
    return it;
  }
  long long cut(long long p, long long id) {
    auto it = get(p);
    auto nit = it;
    nit++;
    if (nit == S.end() || nit->first != p + 1)
      S.insert(pair<long long, long long>(p + 1, id));
    return it->second;
  }
};
long long n, k;
vector<pair<long long, pair<long long, long long> > > qvec;
ArraybySet as;
long long cnt[100005], ans[100005];
long long M = 1000000005;
long long a[400005];
int main(void) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  long long x, y;
  for (long long(i) = (1); (i) <= (n); (i)++) {
    cin >> x >> y;
    qvec.push_back(pair<long long, pair<long long, long long> >(
        y - k + 1, pair<long long, long long>(x - k + 1, x + 1)));
    qvec.push_back(pair<long long, pair<long long, long long> >(
        y + 1, pair<long long, long long>(x + 1, x - k + 1)));
  }
  sort((qvec).begin(), (qvec).end());
  long long id = 1;
  for (auto e : qvec) {
    long long l = e.second.first, r = e.second.second, x = 1;
    if (l > r) swap(l, r), x = -1;
    r--;
    a[id] = a[as.cut(l - 1, id)], id++;
    a[id] = a[as.cut(r, id)], id++;
    for (auto it = as.get(l); it->first <= r;) {
      long long id = it->second, v = a[id], len = -it->first, L = it->first;
      it++, len += it->first;
      a[id] += x;
      ans[v + x] += len * (M - e.first + 1), ans[v] -= len * (M - e.first + 1);
    }
  }
  outa(ans, 1, n);
  return 0;
}

### Prompt

Please formulate a cpp solution to the following problem:

Snow Queen told Kay to form a word "eternity" using pieces of ice. Kay is eager to deal with the task, because he will then become free, and Snow Queen will give him all the world and a pair of skates.

Behind the palace of the Snow Queen there is an infinite field consisting of cells. There are n pieces of ice spread over the field, each piece occupying exactly one cell and no two pieces occupying the same cell. To estimate the difficulty of the task Kay looks at some squares of size k × k cells, with corners located at the corners of the cells and sides parallel to coordinate axis and counts the number of pieces of the ice inside them.

This method gives an estimation of the difficulty of some part of the field. However, Kay also wants to estimate the total difficulty, so he came up with the following criteria: for each x (1 ≤ x ≤ n) he wants to count the number of squares of size k × k, such that there are exactly x pieces of the ice inside.

Please, help Kay estimate the difficulty of the task given by the Snow Queen.

Input

The first line of the input contains two integers n and k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ 300) — the number of pieces of the ice and the value k, respectively. Each of the next n lines contains two integers xi and yi ( - 109 ≤ xi, yi ≤ 109) — coordinates of the cell containing i-th piece of the ice. It's guaranteed, that no two pieces of the ice occupy the same cell.

Output

Print n integers: the number of squares of size k × k containing exactly 1, 2, ..., n pieces of the ice.

Example

Input

5 3
4 5
4 6
5 5
5 6
7 7


Output

10 8 1 4 0 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
struct edge {
  long long to, cost;
  edge() {}
  edge(long long a, long long b) { to = a, cost = b; }
};
const long long dx[] = {1, 0, -1, 0}, dy[] = {0, -1, 0, 1};
const long long mod = 1000000007;
struct mint {
  long long x = 0;
  mint(long long y = 0) {
    x = y;
    if (x < 0 || x >= mod) x = (x % mod + mod) % mod;
  }
  mint(const mint& ope) { x = ope.x; }
  mint operator-() { return mint(-x); }
  mint operator+(const mint& ope) { return mint(x) += ope; }
  mint operator-(const mint& ope) { return mint(x) -= ope; }
  mint operator*(const mint& ope) { return mint(x) *= ope; }
  mint operator/(const mint& ope) { return mint(x) /= ope; }
  mint& operator+=(const mint& ope) {
    x += ope.x;
    if (x >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint& ope) {
    x += mod - ope.x;
    if (x >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint& ope) {
    x *= ope.x, x %= mod;
    return *this;
  }
  mint& operator/=(const mint& ope) {
    long long n = mod - 2;
    mint mul = ope;
    while (n) {
      if (n & 1) *this *= mul;
      mul *= mul;
      n >>= 1;
    }
    return *this;
  }
  mint inverse() { return mint(1) / *this; }
  bool operator==(const mint& ope) { return x == ope.x; }
  bool operator!=(const mint& ope) { return x != ope.x; }
};
mint modpow(mint a, long long n) {
  if (n == 0) return mint(1);
  if (n % 2)
    return a * modpow(a, n - 1);
  else
    return modpow(a * a, n / 2);
}
istream& operator>>(istream& is, mint& ope) {
  long long t;
  is >> t, ope.x = t;
  return is;
}
ostream& operator<<(ostream& os, mint& ope) { return os << ope.x; }
ostream& operator<<(ostream& os, const mint& ope) { return os << ope.x; }
bool exceed(long long x, long long y, long long m) { return x >= m / y + 1; }
void mark() { cout << "*" << endl; }
void yes() { cout << "YES" << endl; }
void no() { cout << "NO" << endl; }
long long sgn(long long x) {
  if (x > 0) return 1;
  if (x < 0) return -1;
  return 0;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long digitnum(long long x, long long b = 10) {
  long long ret = 0;
  for (; x; x /= b) ret++;
  return ret;
}
long long digitsum(long long x, long long b = 10) {
  long long ret = 0;
  for (; x; x /= b) ret += x % b;
  return ret;
}
string lltos(long long x) {
  string ret;
  for (; x; x /= 10) ret += x % 10 + '0';
  reverse(ret.begin(), ret.end());
  return ret;
}
long long stoll(string& s) {
  long long ret = 0;
  for (auto c : s) ret *= 10, ret += c - '0';
  return ret;
}
template <typename T>
void uniq(T& vec) {
  sort(vec.begin(), vec.end());
  vec.erase(unique(vec.begin(), vec.end()), vec.end());
}
template <typename T>
ostream& operator<<(ostream& os, vector<T>& vec) {
  for (int i = 0; i < vec.size(); i++)
    os << vec[i] << (i + 1 == vec.size() ? "" : " ");
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, deque<T>& deq) {
  for (int i = 0; i < deq.size(); i++)
    os << deq[i] << (i + 1 == deq.size() ? "" : " ");
  return os;
}
template <typename T, typename U>
ostream& operator<<(ostream& os, pair<T, U>& pair_var) {
  os << "(" << pair_var.first << ", " << pair_var.second << ")";
  return os;
}
template <typename T, typename U>
ostream& operator<<(ostream& os, const pair<T, U>& pair_var) {
  os << "(" << pair_var.first << ", " << pair_var.second << ")";
  return os;
}
template <typename T, typename U>
ostream& operator<<(ostream& os, map<T, U>& map_var) {
  for (typename map<T, U>::iterator itr = map_var.begin(); itr != map_var.end();
       itr++) {
    os << "(" << itr->first << ", " << itr->second << ")";
    itr++;
    if (itr != map_var.end()) os << ",";
    itr--;
  }
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, set<T>& set_var) {
  for (typename set<T>::iterator itr = set_var.begin(); itr != set_var.end();
       itr++) {
    os << *itr;
    ++itr;
    if (itr != set_var.end()) os << " ";
    itr--;
  }
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, multiset<T>& set_var) {
  for (typename multiset<T>::iterator itr = set_var.begin();
       itr != set_var.end(); itr++) {
    os << *itr;
    ++itr;
    if (itr != set_var.end()) os << " ";
    itr--;
  }
  return os;
}
template <typename T>
void outa(T a[], long long s, long long t) {
  for (long long i = s; i <= t; i++) {
    cout << a[i];
    if (i < t) cout << " ";
  }
  cout << endl;
}
void dump_func() { cout << endl; }
template <class Head, class... Tail>
void dump_func(Head&& head, Tail&&... tail) {
  cout << head;
  if (sizeof...(Tail) > 0) cout << " ";
  dump_func(std::move(tail)...);
}
struct ArraybySet {
  set<pair<long long, long long> > S;
  ArraybySet() { init(); }
  void init() {
    S.clear();
    S.insert(pair<long long, long long>(-1e18, 0));
  }
  set<pair<long long, long long> >::iterator get(long long p) {
    auto it = S.upper_bound(pair<long long, long long>(p, 1e18));
    it--;
    return it;
  }
  long long cut(long long p, long long id) {
    auto it = get(p);
    auto nit = it;
    nit++;
    if (nit == S.end() || nit->first != p + 1)
      S.insert(pair<long long, long long>(p + 1, id));
    return it->second;
  }
};
long long n, k;
vector<pair<long long, pair<long long, long long> > > qvec;
ArraybySet as;
long long cnt[100005], ans[100005];
long long M = 1000000005;
long long a[400005];
int main(void) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  long long x, y;
  for (long long(i) = (1); (i) <= (n); (i)++) {
    cin >> x >> y;
    qvec.push_back(pair<long long, pair<long long, long long> >(
        y - k + 1, pair<long long, long long>(x - k + 1, x + 1)));
    qvec.push_back(pair<long long, pair<long long, long long> >(
        y + 1, pair<long long, long long>(x + 1, x - k + 1)));
  }
  sort((qvec).begin(), (qvec).end());
  long long id = 1;
  for (auto e : qvec) {
    long long l = e.second.first, r = e.second.second, x = 1;
    if (l > r) swap(l, r), x = -1;
    r--;
    a[id] = a[as.cut(l - 1, id)], id++;
    a[id] = a[as.cut(r, id)], id++;
    for (auto it = as.get(l); it->first <= r;) {
      long long id = it->second, v = a[id], len = -it->first, L = it->first;
      it++, len += it->first;
      a[id] += x;
      ans[v + x] += len * (M - e.first + 1), ans[v] -= len * (M - e.first + 1);
    }
  }
  outa(ans, 1, n);
  return 0;
}

```