#include<bits/stdc++.h>
#define IL inline
#define FOR(i, s, t) for(int i = (s); i <= (t); ++i)
#define ROF(i, s, t) for(int i = (s); i >= (t); --i)
#define pii pair<int, int>
#define mkp make_pair
#define fi first
#define se second
#define pb push_back
#define ls (x << 1)
#define rs ((x << 1) | 1)
#define mid ((l + r) >> 1)

using namespace std;

const int N = 5e5 + 5;

struct xld {
	int l, r, d, be, ed;
};
struct req {
	int l, r, k;
};
struct df {
	int lx, a, b, c, id;
};

int n, Q;
vector<int> V[N];
int dfn[N], dep[N], cc = 0;
int Ansn[N];
xld st[N], dd[N << 2];
req que[N];
df so[N << 2];
int nw = 0, tot = 0;

IL int chkmn(int x, int y) {
	return (dep[x] < dep[y] ? x : y);
}

struct seg {
	int t[N << 2];
	IL void pp(int x) {return t[x] = max(t[ls], t[rs]), void();}
	IL void build(int x, int l, int r) {
		if(l == r) return t[x] = 0, void();
		build(ls, l, mid), build(rs, mid + 1, r); 
		return pp(x);
	}
	IL void update(int x, int l, int r, int p, int k) {
		if(l == r) return t[x] = k, void();
		if(p <= mid) update(ls, l, mid, p, k);
		else update(rs, mid + 1, r, p, k); 
		return pp(x);
	}
	IL int query(int x, int l, int r, int a, int b) {
		if(a <= l && b >= r) return t[x];
		int res = 0;
		if(a <= mid) res = max(res, query(ls, l, mid, a, b));
		if(b > mid) res = max(res, query(rs, mid + 1, r, a, b));
		return res;
	}
}s;

struct LC {
	int ST[20][N], Lg[N];
	IL void prework() {
		FOR(i, 2, N - 5) Lg[i] = Lg[i >> 1] + 1;
		return;
	}
	IL void build() {
		prework();
		FOR(j, 1, 19) FOR(i, 1, n - (1 << j) + 1) ST[j][i] = chkmn(ST[j - 1][i], ST[j - 1][i + (1 << (j - 1))]);
		return;
	}
	IL int query(int l, int r) {
		int t = Lg[r - l + 1];
		return chkmn(ST[t][l], ST[t][r - (1 << t) + 1]);
	}
	IL int LCA(int x, int y) {
		if(x == y) return x;
		if((x = dfn[x]) > (y = dfn[y])) swap(x, y);
		return query(x + 1, y);
	}
}L;

IL void dfs(int x, int fa) {
	dfn[x] = ++cc; L.ST[0][cc] = fa; dep[x] = dep[fa] + 1;
	for(int v : V[x]) if(v != fa) dfs(v, x);
	return;
}

int main() {
	freopen("query.in", "r", stdin);
	freopen("query.out", "w", stdout);
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	cin >> n;
	FOR(i, 1, n - 1) {
		int u, v; cin >> u >> v;
		V[u].pb(v), V[v].pb(u);
	} dfs(1, 0); L.build();
	st[++nw] = (xld){1, 1, 1, 1, 0};
	FOR(i, 2, n) {
		int l = 1, r = nw, midn, ansn = 0;
		while(l <= r) {
			midn = (l + r) >> 1; int nd = st[midn].d, lc = L.LCA(nd, i);
			if(chkmn(nd, lc) == lc && lc != nd) ansn = midn, r = midn - 1;
			else l = midn + 1;
		}
		if(ansn == 0) {
			st[++nw] = (xld){i, i, i, i, 0};
			continue;
		} int ll = 0, rr = i - 1, lc = i;
		while(nw >= ansn) st[nw].ed = i - 1, ll = st[nw].l, lc = L.LCA(st[nw].d, i), dd[++tot] = st[nw--];
		st[++nw] = (xld){ll, rr, lc, i, 0}; st[++nw] = (xld){i, i, i, i, 0};
	}
	while(nw) st[nw].ed = n, dd[++tot] = st[nw--]; cin >> Q;
	FOR(i, 1, Q) {
		int l, r, k; cin >> l >> r >> k;
		que[i] = (req){l, r, k}; so[i] = (df){1, l, l + k - 1, 0, i};
	}
	FOR(i, 1, tot) so[i + Q] = (df){0, dd[i].l, dd[i].ed, 0, dep[dd[i].d]}; s.build(1, 1, n);
	sort(so + 1, so + Q + tot + 1, [&](df A, df B){return (A.a == B.a ? A.lx < B.lx : A.a < B.a);});
	FOR(i, 1, Q + tot) {
		int lx = so[i].lx, b = so[i].b, id = so[i].id;
		if(!lx) s.update(1, 1, n, b, id);
		else Ansn[id] = max(Ansn[id], s.query(1, 1, n, b, n));
	} 
	FOR(i, 1, Q) {
		int l = que[i].l, r = que[i].r, k = que[i].k;
		so[i] = (df){1, r - k + 1, r, 0, i};
	}
	FOR(i, 1, tot) so[i + Q] = (df){0, dd[i].l, dd[i].ed, 0, dep[dd[i].d]}; s.build(1, 1, n);
	sort(so + 1, so + Q + tot + 1, [&](df A, df B){return (A.a == B.a ? A.lx < B.lx : A.a < B.a);});
	FOR(i, 1, Q + tot) {
		int lx = so[i].lx, b = so[i].b, id = so[i].id;
		if(!lx) s.update(1, 1, n, b, id);
		else Ansn[id] = max(Ansn[id], s.query(1, 1, n, b, n));
	} s.build(1, 1, n);
	FOR(i, 1, Q) {
		int l = que[i].l, r = que[i].r, k = que[i].k;
		so[i] = (df){1, k - 1, l + k - 1, r, i};
	}
	FOR(i, 1, tot) so[i + Q] = (df){0, dd[i].ed - dd[i].l, max(dd[i].l, dd[i].ed), 0, dep[dd[i].d]};
	sort(so + 1, so + Q + tot + 1, [&](df A, df B){return (A.a == B.a ? A.lx < B.lx : A.a > B.a);});
	FOR(i, 1, Q + tot) {
		int lx = so[i].lx, b = so[i].b, c = so[i].c, id = so[i].id;
		if(!lx) s.update(1, 1, n, b, id);
		else Ansn[id] = max(Ansn[id], s.query(1, 1, n, b, c));
	}
	FOR(i, 1, Q) cout << Ansn[i] << '\n';
	return 0;
}