import sys,bisect,threading
input=sys.stdin.readline
sys.setrecursionlimit(3*(10**5))
def dfs(ans,a,b,v,tre,pb):
for i in range(len(tre[v][0])):
vtnxt=tre[v][0][i]
x=tre[vtnxt][1]
y=tre[vtnxt][2]
a+=x
b+=y
pb.append(b)
c=bisect.bisect_right(pb,a)
ans[vtnxt-1]=c-1
dfs(ans,a,b,vtnxt,tre,pb)
a-=x
b-=y
pb.pop()
def main():
for _ in range(int(input())):
n=int(input())
tre={}
for i in range(1,n+1):
tre[i]=[[],0,0]
for i in range(2,n+1):
p,a,b=map(int,input().split())
tre[p][0].append(i)
tre[i][1]=a
tre[i][2]=b
ans=[0]*n
a=0
b=0
pb=[0]
dfs(ans,a,b,1,tre,pb)
print(*ans[1:])
threading.stack_size(10 ** 8)
t = threading.Thread(target=main)
t.start()
t.join()
#include <bits/stdc++.h>
using namespace std;
#define ios ios::sync_with_stdio(0)
#define endl '\n'
#define int long long
#define ar array<int, 2>
#define arr array<int, 3>
const int N = 2e5 + 1000, M = 2 * N;
const int inf = 0x3f3f3f3f;
int mod = 998244353; //1e9+7;
int t, n, m, k;
int sum[N];
int a[N], b[N];
int ans[N];
map<int, vector<int>> mp;
void dfs(int u, int p, int d)
{
sum[d] = sum[d - 1] + b[u];
a[u] += a[p];
int l = upper_bound(sum + 1, sum + d + 1, a[u]) - sum - 2;
ans[u] = l;
d++;
for (int v : mp[u])
dfs(v, u, d);
};
signed main()
{
ios;
#ifdef DEBUG
freopen("../1.in", "r", stdin);
#endif
// 就是左边的固定长度。。有点最多到哪里。。还比左边的短。。或者等于。
// 这题就是树上回溯+二分的一个运用题。。
//
cin >> t;
while (t--)
{
mp.clear();
cin >> n;
for (int i = 2; i <= n; ++i)
{
// 他没有方向的。。草。。。麻烦。
// you d ..
int x;
cin >> x >> a[i] >> b[i];
mp[x].push_back(i);
}
dfs(1, 0, 1);
for (int i = 2; i <= n; ++i)
cout << ans[i] << ' ';
cout << endl;
}
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