Educational Codeforces Round 10 D. Nested Segments

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本文深入探讨了一个涉及复杂数据结构问题的代码实现，包括时间限制、内存限制、输入输出规范等关键要素。通过离散化和线段树技术优化了空间和时间复杂度，实现了高效的数据处理。实例演示了如何在限定条件下准确计算每个区间包含的区间数量，为解决类似问题提供了一种有效的方法论。

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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 l_i and r_i ( - 10⁹ ≤ l_i < r_i ≤ 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 a_j — the number of segments contained in the j-th segment.

Examples

Input

Output

Input

Output

0
1
1

tips : 注意离散化就行，用右端点的id号来替换右端点的值，使1e9的空间降到1e5，之后用线段树来保存，降低时间复杂度；

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>

using namespace std;
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
const int maxn = 200005;
int SUM[maxn<<3];

int max(int a,int b){if(a>b)return a;else return b;}
int min(int a,int b){if(a<b)return a;else return b;}


struct node
{
    int x;
    int y;
    int id;
}data[200005];


int cmp(node a, node b)
{
    if(a.x>b.x)
        return 1;
    else if(a.x==b.x)
    {
        return a.y>b.y;
    }
    return 0;
}

int cmp1(node a,node b)
{
    if(a.y==b.y)
        return a.x<b.x;
    return a.y<b.y;
}


void PushUP(int rt)
{
  SUM[rt] = SUM[rt<<1] + SUM[rt<<1|1];
}

void build(int l,int r,int rt) {
  if (l == r)
  {
    SUM[rt] = 0;
    //printf("mi = %d\n",MIN[rt]);
    //printf("ma = %d\n",MAX[rt]);
    return ;
  }
  int m = (l + r) >> 1;
  build(lson);
  build(rson);
  PushUP(rt);
}
void update1(int p,int add,int l,int r,int rt)
{
  if (l == r) {
    SUM[rt] = SUM[rt] + add;
    return ;
  }
  int m = (l + r) >> 1;
  if (p <= m) update1(p , add ,lson);
  else update1(p , add , rson);
  PushUP(rt);
}

int queryhe(int L,int R,int l,int r,int rt)
{
  if (L <= l && r <= R)
  {
    return SUM[rt];
  }
  int m = (l + r) >> 1;
  int ret = 0;
  if (L <= m) ret += queryhe(L , R , lson);
  if (R > m)  ret +=  queryhe(L , R , rson);
  return ret;
}


int ans[200005];


int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        int i;

        for(i=1;i<=n;i++)
        {
            scanf("%d%d",&data[i].x,&data[i].y);
            data[i].id=i;
        }

        build(1, n, 1);

        sort(data+1,data+1+n,cmp1);

        for(i=1;i<=n;i++)
        {
            data[i].y=i;
        }

        sort(data+1,data+1+n,cmp);
        for(i=1;i<=n;i++)
        {
           int xu=data[i].id;
           ans[xu]=queryhe(1,data[i].y,1,n,1);
           update1(data[i].y,1,1,n,1);
        }

        for(i=1;i<=n;i++)
            printf("%d\n",ans[i]);

    }
    return 0;
}