You are given n segments on a coordinate line; each endpoint of every segment has integer coordinates. Some segments can degenerate to points. Segments can intersect with each other, be nested in each other or even coincide.

Your task is the following: for every k ∈ [1..n], calculate the number of points with integer coordinates such that the number of segments that cover these points equals k. A segment with endpoints l_i and r_i covers point x if and only if l_i ≤ x ≤ r_i.

Input

The first line of the input contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of segments.

The next n lines contain segments. The i-th line contains a pair of integers l_i, r_i (0 ≤ l_i ≤ r_i ≤ 10^{18}) — the endpoints of the i-th segment.

Output

Print n space separated integers cnt_1, cnt_2, ..., cnt_n, where cnt_i is equal to the number of points such that the number of segments that cover these points equals to i.

Examples

Input

3
0 3
1 3
3 8


Output

6 2 1 


Input

3
1 3
2 4
5 7


Output

5 2 0 

Note

The picture describing the first example:

<image>

Points with coordinates [0, 4, 5, 6, 7, 8] are covered by one segment, points [1, 2] are covered by two segments and point [3] is covered by three segments.

The picture describing the second example:

<image>

Points [1, 4, 5, 6, 7] are covered by one segment, points [2, 3] are covered by two segments and there are no points covered by three segments.
#include <bits/stdc++.h>
using namespace std;
void readi(int &x) {
  int v = 0, f = 1;
  char c = getchar();
  while (!isdigit(c) && c != '-') c = getchar();
  if (c == '-')
    f = -1;
  else
    v = v * 10 + c - '0';
  while (isdigit(c = getchar())) v = v * 10 + c - '0';
  x = v * f;
}
void readll(long long &x) {
  long long v = 0ll, f = 1ll;
  char c = getchar();
  while (!isdigit(c) && c != '-') c = getchar();
  if (c == '-')
    f = -1;
  else
    v = v * 10 + c - '0';
  while (isdigit(c = getchar())) v = v * 10 + c - '0';
  x = v * f;
}
void readc(char &x) {
  char c;
  while ((c = getchar()) == ' ')
    ;
  x = c;
}
void writes(string s) { puts(s.c_str()); }
void writeln() { writes(""); }
void writei(int x) {
  if (x < 0) {
    putchar('-');
    x = abs(x);
  }
  if (!x) putchar('0');
  char a[25];
  int top = 0;
  while (x) {
    a[++top] = (x % 10) + '0';
    x /= 10;
  }
  while (top) {
    putchar(a[top]);
    top--;
  }
}
void writell(long long x) {
  if (x < 0) {
    putchar('-');
    x = abs(x);
  }
  if (!x) putchar('0');
  char a[25];
  int top = 0;
  while (x) {
    a[++top] = (x % 10) + '0';
    x /= 10;
  }
  while (top) {
    putchar(a[top]);
    top--;
  }
}
inline long long inc(int &x) { return ++x; }
inline long long inc(long long &x) { return ++x; }
inline long long inc(int &x, long long y) { return x += y; }
inline long long inc(long long &x, long long y) { return x += y; }
inline double inc(double &x, double y) { return x += y; }
inline long long dec(int &x) { return --x; }
inline long long dec(long long &x) { return --x; }
inline long long dec(int &x, long long y) { return x -= y; }
inline long long dec(long long &x, long long y) { return x -= y; }
inline double dec(double &x, double y) { return x -= y; }
inline long long mul(int &x) { return x = ((long long)x) * x; }
inline long long mul(long long &x) { return x = x * x; }
inline long long mul(int &x, long long y) { return x *= y; }
inline long long mul(long long &x, long long y) { return x *= y; }
inline double mul(double &x, double y) { return x *= y; }
inline long long divi(const int &x) {
  long long ans, l, r, mid;
  ans = 0;
  l = 0;
  r = 0x3fffffff;
  while (l < r) {
    mid = (l + r) / 2;
    if (mid * mid <= x) {
      ans = mid;
      l = mid + 1;
    } else
      r = mid;
  }
  return ans;
}
inline long long divi(const long long &x) {
  long long ans, l, r, mid;
  ans = 0;
  l = 0;
  r = 0x3fffffff;
  while (l < r) {
    mid = (l + r) / 2;
    if (mid * mid <= x) {
      ans = mid;
      l = mid + 1;
    } else
      r = mid;
  }
  return ans;
}
inline long long divi(int &x, long long y) { return x /= y; }
inline long long divi(long long &x, long long y) { return x /= y; }
inline double divi(double &x, double y) { return x /= y; }
inline long long mod(int &x, long long y) { return x %= y; }
inline long long mod(long long &x, long long y) { return x %= y; }
long long n, m, i, j, l, r, ans[200005], f, lst;
map<long long, long long> cnt;
int main() {
  readll(n);
  for (i = 1; i <= n; i++) {
    readll(l);
    readll(r);
    cnt[l]++;
    cnt[r + 1]--;
  }
  lst = f = 0;
  for (__typeof((cnt).begin()) it = (cnt).begin(); it != (cnt).end(); it++) {
    ans[f] += (it->first - lst);
    lst = it->first;
    f += it->second;
  }
  if ((1) <= ((n)))
    for (((i)) = (1); ((i)) <= ((n)); ((i))++) printf("%I64d ", ans[i]);
  return 0;
}
