Welcoming autumn evening is the best for walking along the boulevard and n people decided to do so.

The boulevard can be represented as the axis Ox. For every person there are three parameters characterizing the behavior: t_i, s_i, f_i — the moment of time when the i-th person starts walking, the start point and the end point of the walk respectively. Each person moves in a straight line along the boulevard from s_i to f_i with a constant speed of either 1 or  - 1 depending on the direction.

When the i-th person appears on the boulevard at the point s_i she immediately starts walking towards the point f_i.

If two or more persons meet at the boulevard (they are at the same point at the same time, no matter which directions they are going) they all greet each other. Like in the normal life, every pair of people greet each other at most once.

You task is to calculate for every person how many people she greets while walking along the boulevard.

Please, pay attention to the fact that i-th person may meet and greet any other person at points s_i and f_i. After a person achieves the destination point f_i she moves out of the boulevard and cannot greet anyone else. The same rule applies to the start of the walk: a person cannot greet anyone until she appears on the boulevard.

Input

In the first line there is an integer n (2 ≤ n ≤ 1000) — the number of people who decided to go for a walk.

The following n lines contain parameters for n people. In the i-th line there are three positive integers t_i, s_i, f_i(1 ≤ t_i, s_i, f_i ≤ 10⁶,  s_i ≠ f_i), where t_i, s_i, f_i — the moment of time when the i-th person starts walking, the start point and the end point of the walk respectively.

Output

The single line of the output should contain a sequence of n integers r₁, r₂, ..., r_n separated by a space, where r_i denotes the number which the i-th person greets other people while walking along the boulevard.

Examples

31 1 105 8 29 9 10

2 1 1 

33 2 44 3 43 6 4

2 2 2 题目大意：输入n,代表有n个人，接下来n行代表每个人的开始时间，出发点，终点，速度都为1或者-1，按照出发点和终点位置判断，两个人见面时要打招呼，前提是一定要开始走动，走完或者没有走都不能打招呼叫你求每个人要打多少次招呼个人思路：比赛的时候没时间做这道题，刚刚补题，没想到直接ac，不过这道题挺简单的，思路都在代码中

#include<iostream>
#include<string.h>
#include<map>
#include<cstdio>
#include<cstring>
#include<stdio.h>
#include<cmath>
#include<math.h>
#include<algorithm>
#include<set>
#include<queue>
typedef long long ll;
using namespace std;
const ll mod=1e9+7;
const int maxn=1e3+10;
const int maxk=100+10;
const int maxx=1e4+10;
const ll maxe=1000+10;
#define INF 0x3f3f3f3f3f3f
#define Lson l,mid,rt<<1
#define Rson mid+1,r,rt<<1|1
struct P
{int time,be,en,sum,di;//分别是开始时间，开始位置，结束位置，打招呼次数，方向
}p[maxn];
int n;
void solve(int i,int j)
{int b1,b2;if(p[i].time<=p[j].time)//第一种情况
    {b1=p[i].be+p[i].di*(p[j].time-p[i].time);//先把初始时间变为一样，先走的先走，后走的不变if((b1>=p[i].be&&b1<=p[i].en)||(b1>=p[i].en&&b1<=p[i].be))//变化之后还在该点范围内
        {int mi=min(abs(p[i].en-b1),abs(p[j].en-p[j].be));//取到达终点时间少的，因为一个到达了就不会再打招呼了if(b1<=p[j].be&&(b1+mi*p[i].di>=p[j].be+mi*p[j].di))//第一个刚开始的位置小于第二个刚开始的位置，变化之后大于了，那么证明肯定有见面的时候
            {p[i].sum++;p[j].sum++;}else if(b1>=p[j].be&&(b1+mi*p[i].di<=p[j].be+mi*p[j].di))//刚开始的位置大于第二个的位置，变化之后小于第二个的位置，证明肯定有见面的时候
            {p[i].sum++;p[j].sum++;}}else//不在范围内，证明第二个点还没有出发就已经到达终点了return ;}else//以下同理
    {b2=p[j].be+p[j].di*(p[i].time-p[j].time);if((b2>=p[j].be&&b2<=p[j].en)||(b2>=p[j].en&&b2<=p[j].be)){int mi=min(abs(p[i].en-p[i].be),abs(p[j].en-b2));if(b2<=p[i].be&&(b2+mi*p[j].di>=p[i].be+mi*p[i].di)){p[i].sum++;p[j].sum++;}else if(b2>=p[i].be&&(b2+mi*p[j].di<=p[i].be+mi*p[i].di)){p[i].sum++;p[j].sum++;}}elsereturn ;}
}
int main()
{cin>>n;for(int i=0;i<n;i++){cin>>p[i].time>>p[i].be>>p[i].en;p[i].sum=0;if(p[i].be<=p[i].en) p[i].di=1;else  p[i].di=-1;}for(int i=0;i<n;i++){for(int j=i+1;j<n;j++){solve(i,j);}}for(int i=0;i<n;i++)cout<<p[i].sum<<" ";cout<<endl;return 0;
}