#include<bits/stdc++.h>
using namespace std;

#define QwQ330AwA return 0
#define ll long long

const int N = 1e5 + 330;
const int V = 1e6 + 6;
int p[V], mu[V];
bool vis[V];
int cnt;
struct Edge {
	int u, v, w;
} e[N];
int now[N];
int L[N], R[N];
int n, m;
vector <int> d[V];

void init(int s) {
	mu[1] = 1;
	for (int i = 2; i <= s; i++) {
		if (!vis[i]) p[++cnt] = i, mu[i] = -1;
		for (int j = 1; j <= cnt && 1ll * i * p[j] <= s; j++) {
			vis[i * p[j]] = 1;
			if (i % p[j] == 0) {
				mu[i * p[j]] = 0;
				break;
			} else mu[i * p[j]] = -mu[i];
		}
	}
}

vector <int> t[N << 2];
int tim[N << 2];
#define ls (p << 1)
#define rs ((p << 1) | 1)
int T;
void upd(int p, int l, int r, int L, int R, int x) {
	if (tim[p] != T) t[p].clear(); 
	tim[p] = T;
	if (L <= l && r <= R) {
		t[p].push_back(x);
		return ;
	}
	int mid = (l + r) / 2;
	if (L <= mid) upd(ls, l, mid, L, R, x);
	if (mid < R) upd(rs, mid + 1, r, L, R, x);
}

ll ans[N];
ll sum;
int siz[N];
int fa[N];
int find(int x) {
	return x == fa[x] ? x : find(fa[x]);
}
int stk[N][2], top;

void merge(int u, int v) {
	u = find(u), v = find(v);
	assert(u != v);
	if (siz[u] < siz[v]) swap(u, v);
	fa[v] = u;
	sum += 1ll * siz[u] * siz[v] * mu[T];
	siz[u] += siz[v];
	top++;
	stk[top][0] = u;
	stk[top][1] = v;
}

void del(int lst) {
	while (top > lst) {
		int u = stk[top][0], v = stk[top][1];
		fa[v] = v;
		siz[u] -= siz[v];
		sum -= 1ll * siz[u] * siz[v] * mu[T];
		top--;
	}
}

void dfs(int p, int l, int r) {
	if (tim[p] != T) {
		ans[l] += sum;
		ans[r + 1] -= sum;
		return ;
	}
	int lst = top;
	for (auto i : t[p]) merge(e[i].u, e[i].v);
	if (l == r) {
		ans[l] += sum;
		ans[r + 1] -= sum;
	} else {
		int mid = (l + r) / 2;
		dfs(ls, l, mid);
		dfs(rs, mid + 1, r);
	}
	del(lst);
}

signed main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	init(1e6);
	cin >> n;
	for (int i = 1, u, v, w; i < n; i++) {
		cin >> u >> v >> w;
		e[i] = {u, v, w};
		now[i] = i;
	}
	cin >> m;
	for (int i = 1; i < n; i++) L[i] = 0, R[i] = m;
	for (int i = 1; i <= m; i++) {
		int x, y;
		cin >> x >> y;
		int id = now[x];
		R[id] = i - 1;
		L[n + i - 1] = i;
		R[n + i - 1] = m;
		e[n + i - 1] = e[id];
		e[n + i - 1].w = y;
		now[x] = n + i - 1;
	}
	for (int i = 1; i <= n + m - 1; i++) {
		int w = e[i].w;
		for (int j = 1; 1ll * j * j <= w; j++) {
			if (w % j == 0) {
				if (mu[j] != 0) d[j].push_back(i);
				if (j * j != w && mu[w / j] != 0) d[w / j].push_back(i);
			} 
		}
	}
	for (int i = 1; i <= n; i++) siz[i] = 1, fa[i] = i;
	for (int i = 1; i <= 1e6; i++) {
		if (mu[i] == 0) continue;
		T = i;
		for (auto id : d[i]) upd(1, 0, m, L[id], R[id], id);
		dfs(1, 0, m);
	}
	for (int i = 1; i <= m; i++) ans[i] += ans[i - 1];
	for (int i = 0; i <= m; i++) cout << ans[i] << '\n';
	QwQ330AwA;
}

/*
  枚举 $d$，开放为 $d$ 的倍数路径，计算连通块贡献即可，乘个 mu
  多测怎么做啊
  我靠就100组
  感觉可以吧1e6*100*log
  不行啊
  直接开V颗LCT每次断边
  线段树分治换可撤销并查集
  总共nlogV条边
  每条边修改logn个结点
  3个log卧槽
  应该跑不满吧
  那这个题q可以开到1e6啊
  哦可以每条边都拍下来然后直接撤销
  那我觉得挺弱智的
  感觉赢了
*/