void cdq(int l, int r){if(l==r) return;int m = (l+r)>>1;cdq(l,m);cdq(m+1,r);//按y进行排序，那么问题就变成两个y递增的集合，//后一个集合中的每个y在前一个集合中有多少个y小于等于它sort(a+l,a+m+1,cmp);sort(a+m+1,a+r+1,cmp);int j = l;for(int i=m+1;i<=r;++i){for(;j<=m;&&a[j].y<=a[i].y;++j);a[i].sum += j - l;}
}

#include <stdio.h>
#include <math.h>
#include <algorithm>
#include <iostream>
#include <string.h>
using namespace std;
struct Point{int x,y,z;int id;int sum;Point(){}Point(int x, int y):x(x),y(y){}bool operator<(const Point&rhs)const{if(x!=rhs.x) return x < rhs.x;if(y!=rhs.y) return y < rhs.y;return z < rhs.z;}bool operator==(const Point &rhs)const{return x==rhs.x && y==rhs.y && z==rhs.z;}
};
bool cmp(const Point &lhs, const Point &rhs){if(lhs.y!=rhs.y) return lhs.y <rhs.y;return lhs.z <rhs.z;
}
const int N = 100000 + 10;
Point a[N];
class BIT{
public:int sum[N];int n;void init(){n = 100000;memset(sum,0,sizeof(sum));}int lowbit(int x){return x & (-x);}int modify(int x, int val){while(x<=n){sum[x] += val;x += lowbit(x);}}int getSum(int x){int ret= 0;while(x>0){ret += sum[x];x -= lowbit(x);}return ret;}
}bit;void cdq(int l, int r){if(l==r)return;int m = (l+r)>>1;cdq(l,m);cdq(m+1,r);sort(a+l,a+m+1,cmp);sort(a+m+1,a+r+1,cmp);int j = l;for(int i=m+1;i<=r;++i){for(;j<=m &&a[j].y<=a[i].y;++j)bit.modify(a[j].z,1);a[i].sum += bit.getSum(a[i].z);}for(int i=l; i<j; ++i)bit.modify(a[i].z,-1);}int ans[N];
int main(){int t,n;scanf("%d",&t);while(t--){scanf("%d",&n);for(int i=0;i<n;++i){ scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].z); a[i].id = i; a[i].sum=0;}sort(a,a+n);bit.init();cdq(0,n-1);sort(a,a+n);for(int i=0;i<n;){int j = i + 1;int tmp = a[i].sum;//分治时，坐标相等的时候，//排在前边的坐标不能使用后边的坐标更新自己，所以要在这里处理一下for(;j<n &&a[i]==a[j];++j) tmp = max(tmp,a[j].sum);for(int k=i;k<j;++k) ans[a[k].id] = tmp;i = j;}for(int i=0;i<n;++i)printf("%d\n",ans[i]);}return 0;
}

#include <stdio.h>
#include <math.h>
#include <algorithm>
#include <iostream>
#include <string.h>
using namespace std;
struct Point{int x,y,z;int id;int sum;Point(){}Point(int x, int y):x(x),y(y){}bool operator<(const Point&rhs)const{if(x!=rhs.x) return x < rhs.x;if(y!=rhs.y) return y < rhs.y;return z < rhs.z;}bool operator==(const Point &rhs)const{return x==rhs.x && y==rhs.y && z==rhs.z;}
};
bool cmp(const Point &lhs, const Point &rhs){if(lhs.y!=rhs.y) return lhs.y <rhs.y;return lhs.z <rhs.z;
}
const int N = 100000 + 10;
Point a[N];
class BIT{
public:int sum[N];int n;void init(){n = 100000;memset(sum,0,sizeof(sum));}int lowbit(int x){return x & (-x);}int modify(int x, int val){while(x<=n){sum[x] += val;x += lowbit(x);}}int getSum(int x){int ret= 0;while(x>0){ret += sum[x];x -= lowbit(x);}return ret;}
}bit;Point tmp[N];
void cdq(int l, int r){if(l==r)return;int m = (l+r)>>1;cdq(l,m);cdq(m+1,r);//sort(a+l,a+m+1,cmp);//sort(a+m+1,a+r+1,cmp);int j = l;for(int i=m+1;i<=r;++i){for(;j<=m &&a[j].y<=a[i].y;++j)bit.modify(a[j].z,1);a[i].sum += bit.getSum(a[i].z);}for(int i=l; i<j; ++i)bit.modify(a[i].z,-1);//归并排序， 这样就不需要上面的sort了int i = l ;j = m+1;for(int k=l;k<=r;++k){if(i>m) tmp[k] = a[j++];else if(j>r) tmp[k] = a[i++];else if(a[i].y < a[j].y) tmp[k] = a[i++];else tmp[k] = a[j++];}for(int k=l;k<=r;++k)a[k] = tmp[k];}int ans[N];
int main(){int t,n;scanf("%d",&t);while(t--){scanf("%d",&n);for(int i=0;i<n;++i){ scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].z); a[i].id = i; a[i].sum=0;}sort(a,a+n);bit.init();cdq(0,n-1);sort(a,a+n);for(int i=0;i<n;){int j = i + 1;int tmp = a[i].sum;//分治时，坐标相等的时候，//排在前边的坐标不能使用后边的坐标更新自己，所以要在这里处理一下for(;j<n &&a[i]==a[j];++j) tmp = max(tmp,a[j].sum);for(int k=i;k<j;++k) ans[a[k].id] = tmp;i = j;}for(int i=0;i<n;++i)printf("%d\n",ans[i]);}return 0;
}