#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXN = 200000;
typedef struct S {
  int l, r, idx;
} S;
bool operator<(const S &a, const S &b) {
  if (a.l != b.l) return a.l > b.l;
  if (a.r != b.r) return a.r < b.r;
  return a.idx < b.idx;
}
int n;
S s[MAXN];
int x[2 * MAXN], nx;
int ret[MAXN];
int bt[2 * MAXN + 1];
int bget(int idx) {
  ++idx;
  int ret = 0;
  while (idx > 0) {
    ret += bt[idx];
    idx -= idx & -idx;
  }
  return ret;
}
void bmod(int idx, int by) {
  ++idx;
  while (idx <= nx) {
    bt[idx] += by;
    idx += idx & -idx;
  }
}
void run() {
  scanf("%d", &n);
  for (int i = (0); i < (n); ++i) scanf("%d%d", &s[i].l, &s[i].r), s[i].idx = i;
  nx = 0;
  for (int i = (0); i < (n); ++i) x[nx++] = s[i].l, x[nx++] = s[i].r;
  sort(x, x + nx);
  nx = unique(x, x + nx) - x;
  for (int i = (0); i < (n); ++i)
    s[i].l = lower_bound(x, x + nx, s[i].l) - x,
    s[i].r = lower_bound(x, x + nx, s[i].r) - x;
  sort(s, s + n);
  memset(bt, 0, sizeof(bt));
  for (int i = (0); i < (n); ++i) {
    ret[s[i].idx] = bget(s[i].r);
    bmod(s[i].r, +1);
  }
  for (int i = (0); i < (n); ++i) printf("%d\n", ret[i]);
}
int main() {
  run();
  return 0;
}

### Prompt

Create a solution in CPP for 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 gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXN = 200000;
typedef struct S {
  int l, r, idx;
} S;
bool operator<(const S &a, const S &b) {
  if (a.l != b.l) return a.l > b.l;
  if (a.r != b.r) return a.r < b.r;
  return a.idx < b.idx;
}
int n;
S s[MAXN];
int x[2 * MAXN], nx;
int ret[MAXN];
int bt[2 * MAXN + 1];
int bget(int idx) {
  ++idx;
  int ret = 0;
  while (idx > 0) {
    ret += bt[idx];
    idx -= idx & -idx;
  }
  return ret;
}
void bmod(int idx, int by) {
  ++idx;
  while (idx <= nx) {
    bt[idx] += by;
    idx += idx & -idx;
  }
}
void run() {
  scanf("%d", &n);
  for (int i = (0); i < (n); ++i) scanf("%d%d", &s[i].l, &s[i].r), s[i].idx = i;
  nx = 0;
  for (int i = (0); i < (n); ++i) x[nx++] = s[i].l, x[nx++] = s[i].r;
  sort(x, x + nx);
  nx = unique(x, x + nx) - x;
  for (int i = (0); i < (n); ++i)
    s[i].l = lower_bound(x, x + nx, s[i].l) - x,
    s[i].r = lower_bound(x, x + nx, s[i].r) - x;
  sort(s, s + n);
  memset(bt, 0, sizeof(bt));
  for (int i = (0); i < (n); ++i) {
    ret[s[i].idx] = bget(s[i].r);
    bmod(s[i].r, +1);
  }
  for (int i = (0); i < (n); ++i) printf("%d\n", ret[i]);
}
int main() {
  run();
  return 0;
}

```