#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define pb push_back
#define pii pair<ll,ll>
#define f first
#define s second
#define chmin(a, b) a = min(a,b)
#define chmax(a, b) a = max(a,b)
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define F0R(i, a) for (int i = 0; i < (a); i++)
#define all(x) x.begin(),x.end()
#define vec vector
const int MAX_N = 8e4+1, l2d = 18;
pair<pii,int> pts[3*MAX_N];
pair<pii,pii> p2[MAX_N];
int lift[MAX_N][l2d];
vec<int> adj[MAX_N];
set<int> col[MAX_N];
int ans[MAX_N], c[MAX_N];
int jump(int a, int d) {
for (int i=0;i<l2d;i++) {
if ((1<<i)&d) {
a = lift[a][i];
if (a==-1) {a = 0;}
}
} return a;
}
int main() {
ios::sync_with_stdio(false); cin.tie(0);
int n, m; cin >> n >> m;
memset(lift,-1,sizeof(lift));
p2[0] = {{0,0},{2e9,2e9}};
for (int i=0;i<n;i++) {
cin >> pts[i].f.f >> pts[i].f.s;
cin >> pts[i+n+m].f.f >> pts[i+n+m].f.s;
pts[i].s = i+1; pts[i+n+m].s = i+n+m+1;
p2[i+1] = {pts[i].f,pts[i+n+m].f};
} // [1,n]: pts, [n+1,n+m] balls, [n+m+1,2*n+m] pts2
for (int i=n;i<n+m;i++) {
cin >> pts[i].f.f >> pts[i].f.s >> c[i-n+1];
pts[i].s = i+1;
}
sort(pts,pts+2*n+m);
set<pii> active;
active.insert({0,0});
for (int i=0;i<2*n+m;i++) {
int idx = pts[i].s;
auto[x,y] = pts[i].f;
if (idx<=n+m) {
int it = prev(active.upper_bound({y,1e9}))->s;
int lo = 0, hi = n, mid = (lo+hi)/2, ans = -1;
int goal = (idx<=n ? p2[idx].s.s : y);
while (lo <= hi) {
int x2 = jump(it,mid);
if (p2[x2].s.s>=goal) {hi = mid-1; ans = x2;}
else {lo = mid+1;}
mid = (lo+hi)/2;
}
if (idx<=n) {
active.insert({y,idx});
adj[ans].pb(idx);
lift[idx][0] = ans;
for (int j=1;j<l2d;j++) {
if (lift[idx][j-1]!=-1) {lift[idx][j] = lift[lift[idx][j-1]][j-1];}
}
} else {col[ans].insert(c[idx-n]);}
} else {
active.erase(active.find({p2[idx-n-m].f.s,idx-n-m}));
}
}
function<void(int,int)> dfs = [&](int cur, int par) {
for (int x : adj[cur]) {
if (x!=par) {
dfs(x,cur);
if (col[x].size()>col[cur].size()) {swap(col[x],col[cur]);}
for (int i : col[x]) {col[cur].insert(i);}
}
} ans[cur] = col[cur].size();
};
dfs(0,0);
for (int i=1;i<=n;i++) {cout << ans[i] << '\n';}
/*
sort points and ranges
process start range, then point, then end range
binary search ez to build tree
get the thing in set and optimal must be an ancestor of it
nlogn; sparse table get max with binary search so (m+n)logn
now calc distinct values
nlog^2m because merging
*/
}
