stars

Astronomers often examine star maps where stars are represented by points on a plane and each star has Cartesian coordinates. Let the level of a star be an amount of the stars that are not higher and not to the right of the given star. Astronomers want to know the distribution of the levels of the stars.

For example, look at the map shown on the figure above. Level of the star number 5 is equal to 3 (it's formed by three stars with a numbers 1, 2 and 4). And the levels of the stars numbered by 2 and 4 are 1. At this map there are only one star of the level 0, two stars of the level 1, one star of the level 2, and one star of the level 3.

You are to write a program that will count the amounts of the stars of each level on a given map.

Input

The first line of the input file contains a number of stars N (1<=N<=15000). The following N lines describe coordinates of stars (two integers X and Y per line separated by a space, 0<=X,Y<=32000). There can be only one star at one point of the plane. Stars are listed in ascending order of Y coordinate. Stars with equal Y coordinates are listed in ascending order of X coordinate.

Output

The output should contain N lines, one number per line. The first line contains amount of stars of the level 0, the second does amount of stars of the level 1 and so on, the last line contains amount of stars of the level N-1.

Sample Input

51 15 17 13 35 5

Sample Output

12110

Hint

This problem has huge input data,use scanf() instead of cin to read data to avoid time limit exceed.

这道题本质仍然是用树状数组处理逆序对数的问题，唯一不同的是，对于每一个数，将其插入序列之后我们可以得到与它形成逆序的数的个数，用该数所在的位置减去这个数即是它的level。

#include<cstdio>
#include<cstring>
#include<algorithm>
#define MAXN 15005
using namespace std;
struct node {
 int data;
 int pos;
}a[MAXN];
int c[MAXN],lev[MAXN];
int n;
bool cmp(node a,node b){
 if(a.data ==b.data ) return a.pos <b.pos ;
 return a.data <b.data ;
}
bool cmp1(node a,node b){
 return a.pos <b.pos ;
}
int lowbit(int x){
 return x&(-x);
}
void update(int x,int val){
     for(;x<=n;x+=lowbit(x))
  c[x]+=val;
}
int getsum(int x){
 int sum=0;
 for(;x;x-=lowbit(x)){
  sum+=c[x];
 }
 return sum;
}
int main(){
 scanf("%d",&n);
 memset(c,0,sizeof(c));
 memset(lev,0,sizeof(lev));
 int k;
 for(int i=1;i<=n;i++){
  scanf("%d %d",&a[i].data ,&k );
  a[i].pos =i;
 }
 sort(a+1,a+n+1,cmp);
 for(int i=1;i<=n;i++){
  a[i].data =i;
 }
 sort(a+1,a+n+1,cmp1);
 int t;
 for(int i=1;i<=n;i++){
  update(a[i].data ,1);
  t=i-getsum(a[i].data );
  lev[i-t]++;
 }
 for(int i=1;i<=n;i++){
  printf("%d\n",lev[i]);
 }
 return 0;
}