#include <bits/stdc++.h>
using namespace std;
long long n, m, p[200005];
long long a[4 * 200005];
pair<long long, long long> val[200005];
void st(long long ind, long long val) {
  while (ind < 200005) {
    a[ind] += val;
    ind += (ind & -ind);
  }
  return;
}
long long gt(long long ind) {
  long long ret = 0;
  while (ind) {
    ret += a[ind];
    ind -= (ind & -ind);
  }
  return ret;
}
bool comp(pair<long long, long long> a, pair<long long, long long> b) {
  return a.second < b.second;
}
map<long long, long long> mp;
map<long long, long long> ord;
long long res[200005];
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> val[i].first >> val[i].second;
    ord[val[i].first] = i;
  }
  sort(val + 1, val + n + 1, comp);
  for (int i = 1; i <= n; i++) {
    mp[val[i].second] = i;
    st(i, 1);
  }
  sort(val + 1, val + n + 1);
  for (int i = 1; i <= n; i++) {
    res[ord[val[i].first]] = gt(mp[val[i].second]) - 1;
    st(mp[val[i].second], -1);
  }
  for (int i = 1; i <= n; i++) {
    cout << res[i] << endl;
  }
  return 0;
}

### Prompt

Please create a solution in cpp to the following problem:

You are given n segments on a line. There are no ends of some segments that coincide. For each segment find the number of segments it contains.

Input

The first line contains a single integer n (1 ≤ n ≤ 2·105) — the number of segments on a line.

Each of the next n lines contains two integers li and ri ( - 109 ≤ li < ri ≤ 109) — the coordinates of the left and the right ends of the i-th segment. It is guaranteed that there are no ends of some segments that coincide.

Output

Print n lines. The j-th of them should contain the only integer aj — the number of segments contained in the j-th segment.

Examples

Input

4
1 8
2 3
4 7
5 6


Output

3
0
1
0


Input

3
3 4
1 5
2 6


Output

0
1
1

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long n, m, p[200005];
long long a[4 * 200005];
pair<long long, long long> val[200005];
void st(long long ind, long long val) {
  while (ind < 200005) {
    a[ind] += val;
    ind += (ind & -ind);
  }
  return;
}
long long gt(long long ind) {
  long long ret = 0;
  while (ind) {
    ret += a[ind];
    ind -= (ind & -ind);
  }
  return ret;
}
bool comp(pair<long long, long long> a, pair<long long, long long> b) {
  return a.second < b.second;
}
map<long long, long long> mp;
map<long long, long long> ord;
long long res[200005];
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> val[i].first >> val[i].second;
    ord[val[i].first] = i;
  }
  sort(val + 1, val + n + 1, comp);
  for (int i = 1; i <= n; i++) {
    mp[val[i].second] = i;
    st(i, 1);
  }
  sort(val + 1, val + n + 1);
  for (int i = 1; i <= n; i++) {
    res[ord[val[i].first]] = gt(mp[val[i].second]) - 1;
    st(mp[val[i].second], -1);
  }
  for (int i = 1; i <= n; i++) {
    cout << res[i] << endl;
  }
  return 0;
}

```