1 #include<cstdio>
2 #include<cstring>
3 #include<vector>
4 #include<algorithm>
5 #include<set>
6 using namespace std;
7 #define MAXN 16010
8 #define INF 123456789
9 struct Seg
10 {
11 int y1,y2,x;
12 friend bool operator<(Seg a,Seg b)
13 {
14 return a.x<b.x;
15 }
16 };
17 int i,tree[MAXN<<2];
18 Seg p[MAXN];
19 vector<int>G[MAXN];
20 vector<int>::iterator it;
21 inline void PushDown(int rt)
22 {
23 if(tree[rt])
24 {
25 tree[rt<<1]=tree[rt<<1|1]=tree[rt];
26 tree[rt]=0;
27 }
28 }
29 void Update(int x,int y,int val,int L,int R,int rt)
30 {
31 if(x<=L&&R<=y)
32 tree[rt]=val;
33 else
34 {
35 int mid=(L+R)>>1;
36 PushDown(rt);
37 if(x<=mid)
38 Update(x,y,val,L,mid,rt<<1);
39 if(y>mid)
40 Update(x,y,val,mid+1,R,rt<<1|1);
41 }
42 }
43 void Query(int x,int y,int L,int R,int rt)
44 {
45 if(x<=L&&R<=y)
46 {
47 if(tree[rt])
48 {
49 G[i].push_back(tree[rt]);
50 G[tree[rt]].push_back(i);
51 }
52 else if(L!=R)
53 {
54 int mid=(L+R)>>1;
55 Query(x,y,L,mid,rt<<1);
56 Query(x,y,mid+1,R,rt<<1|1);
57 }
58 }
59 else
60 {
61 int mid=(L+R)>>1;
62 PushDown(rt);
63 if(x<=mid)
64 Query(x,y,L,mid,rt<<1);
65 if(y>mid)
66 Query(x,y,mid+1,R,rt<<1|1);
67 }
68 }
69 int main()
70 {
71 int t,n,j,k,r,left,right,ans;
72 scanf("%d",&t);
73 while(t--)
74 {
75 left=INF;
76 ans=right=0;
77 scanf("%d",&n);
78 for(i=1;i<=n;i++)
79 {
80 scanf("%d%d%d",&p[i].y1,&p[i].y2,&p[i].x);
81 p[i].y1<<=1;
82 p[i].y2<<=1;
83 left=min(left,p[i].y1);
84 right=max(right,p[i].y2);
85 G[i].clear();
86 }
87 sort(p+1,p+1+n);
88 memset(tree,0,sizeof(tree));
89 for(i=1;i<=n;i++)
90 {
91 Query(p[i].y1,p[i].y2,left,right,1);
92 Update(p[i].y1,p[i].y2,i,left,right,1);
93 }
94 for(i=1;i<=n;i++)
95 {
96 sort(G[i].begin(),G[i].end());
97 it=unique(G[i].begin(),G[i].end());
98 G[i].resize(it-G[i].begin());
99 }
100 for(i=1;i<=n;i++)
101 {
102 for(j=0;j<G[i].size();j++)
103 {
104 for(k=j+1;k<G[i].size();k++)
105 {
106 r=lower_bound(G[G[i][j]].begin(),G[G[i][j]].end(),G[i][k])-G[G[i][j]].begin();
107 if(r<G[G[i][j]].size()&&G[G[i][j]][r]==G[i][k])
108 ans++;
109 }
110 }
111 }
112 printf("%d\n",ans/3);
113 }
114 return 0;
115 }
