#include <bits/stdc++.h> 


using namespace std;
typedef long long ll;
const int N = 1e6 + 5;

int n, m, tot;
int dep[N], a[N], las[N];
int g[N][20]/*指向下一层的指针*/;
ll	s[N][20]/* {[( i 往下的 2 ^ j 层中第 j 位为 1 的数量 )]} 的后缀和*/
	, h[N][20]/*若干层的节点数目*/
	, f[N][20]/*dp*/;
int id[21], len[21];


vector<int> lay[N]/*layer*/, E[N]/*edge*/;

void dfs(int x) {
	for (int y: E[x]) dep[y] = dep[x] + 1, dfs(y);
	if (lay[dep[x] + 1].size()) g[x][0] = lay[dep[x] + 1].back();
	
	if (lay[dep[x]].size()) las[x] = lay[dep[x]].back();
	
	h[x][0] = h[las[x]][0] + 1, lay[dep[x]].push_back(x);
	
	for (int i = 1; i < 20; i++) g[x][i] = g[g[x][i - 1]][i - 1];
	for (int i = 1; i < 20; i++) h[x][i] = h[x][i - 1] + h[g[x][i - 1]][i - 1];
	for (int i = 0; i < 20; i++) 
		s[x][i] = s[g[x][0]][i] + s[las[x]][i] - s[g[las[x]][0]][i] + (a[x] >> i & 1);
	f[x][0] = f[las[x]][0] + a[x];
	for (int i = 1; i < 20; i++) 
		f[x][i] = f[x][i - 1] + f[g[x][i - 1]][i - 1]
		 + h[g[x][i - 1]][i - 1] * (1 << (i - 1))
		 - ((1 << (i - 1)) << 1) * (s[g[x][i - 1]][i - 1] - s[g[x][i]][i - 1]);
	// 1 -> -1 减两次 
}
ll ask(int x, int d) {
	d++;
	id[tot = 0] = x;
	ll ans = 0, sum = 0;
	for (int i = 19; i >= 0 && x; i--) {
		if (d >> i & 1) id[++tot] = x = g[x][i], len[tot] = i;
	}
	
	for (int i = tot; i >= 1; i--) {
//		printf("///// %d %d\n", id[i - 1], len[i]);
		ans += f[id[i - 1]][len[i]]
		 + (1 << len[i]) * sum - 
		((1 << len[i]) << 1) * (s[id[i]][len[i]] - s[id[tot]][len[i]]);
		sum += h[id[i - 1]][len[i]]
		;
	}
//	printf("///////// %d\n", id[tot]);
	return ans;
}

void infor(int x) {
	printf("infor %d:", x);
	printf("\n	las = %d\n", las[x]);
	printf("\n	g: ");
	for (int i = 0; i < 3; i++) {
		printf("%d ", g[x][i]);
	}
	printf("\n	f: ");
	for (int i = 0; i < 3; i++) {
		printf("%lld ", f[x][i]);
	}
	printf("\n	h: ");
	for (int i = 0; i < 3; i++) {
		printf("%lld ", h[x][i]);
	}
	printf("\n	s: ");
	for (int i = 0; i < 3; i++) {
		printf("%lld ", s[x][i]);
	}
	printf("\n");
}

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
	for (int fa, i = 2; i <= n; i++) scanf("%d", &fa), E[fa].push_back(i);
	
	dep[1] = 1, dfs(1);
	
	scanf("%d", &m);
	
//	for (int i = 1; i <= n; i++) infor(i);
	
	for (int x, y, d, i = 1; i <= m; i++) {
		scanf("%d%d", &x, &d);
		y = las[x];
		if (dep[x] != dep[y]) printf("%lld\n", ask(x, d));
		else printf("%lld\n", ask(x, d) - ask(y, d));
	} 
	
	return 0;
}