// Problem: C - Made Up
// Contest: AtCoder - AISing Programming Contest 2021（AtCoder Beginner Contest 202）
// URL: https://atcoder.jp/contests/abc202/tasks/abc202_c
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)
#pragma GCC optimize(2)
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll>PLL;
typedef pair<int, int>PII;
typedef pair<double, double>PDD;
#define I_int ll
inline ll read()
{
    ll x = 0, f = 1;
    char ch = getchar();
    while(ch < ‘0‘ || ch > ‘9‘)
    {
        if(ch == ‘-‘)f = -1;
        ch = getchar();
    }
    while(ch >= ‘0‘ && ch <= ‘9‘)
    {
        x = x * 10 + ch - ‘0‘;
        ch = getchar();
    }
    return x * f;
}
 
inline void out_(ll x){
    if (x < 0) x = ~x + 1, putchar(‘-‘);
    if (x > 9) out_(x / 10);
    putchar(x % 10 + ‘0‘);
}
 
inline void write(ll x){
    if (x < 0) x = ~x + 1, putchar(‘-‘);
    if (x > 9) write(x / 10);
    putchar(x % 10 + ‘0‘);
    puts("");
}
 
#define read read()
#define closeSync ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define multiCase int T;cin>>T;for(int t=1;t<=T;t++)
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i<(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define perr(i,a,b) for(int i=(a);i>(b);i--)
ll ksm(ll a, ll b, ll p)
{
    ll res = 1;
    while(b)
    {
        if(b & 1)res = res * a % p;
        a = a * a % p;
        b >>= 1;
    }
    return res;
}
const int inf = 0x3f3f3f3f;
#define PI acos(-1)
const int maxn=2e5+100;
 
int n,q;
vector<int>g[maxn];
vector<int>dis[maxn];
int in[maxn],out[maxn],tot;
 
void dfs(int u,int fa,int tmpd){
    in[u]=++tot;
    dis[tmpd].push_back(in[u]);
    for(int i=0;i<g[u].size();i++){
        int j=g[u][i];
        if(j==fa) continue;
        dfs(j,u,tmpd+1);
    }
    out[u]=tot;
}
 
int main(){
    n=read;
    rep(i,2,n){
        int x=read;
        g[x].push_back(i);
        g[i].push_back(x);      
    }
    dfs(1,-1,0);
    /*rep(i,1,n){
        cout<<i<<"*******"<<in[i]<<"******"<<out[i]<<"\n";
    }
    for(int i=0;i<=4;i++){
        for(int j=0;j<dis[i].size();j++){
            cout<<dis[i][j]<<"=====";
        }
        puts("");
    }
    */
    q=read;
    while(q--){
        int u=read,d=read;
        int res=upper_bound(dis[d].begin(),dis[d].end(),out[u])-dis[d].begin();
        int ans=lower_bound(dis[d].begin(),dis[d].end(),in[u])-dis[d].begin();
        //cout<<res-ans<<endl;
        printf("%d\n",res-ans);
    }
    return 0;
} 
 

AISing Programming Contest 2021（AtCoder Beginner Contest 202）E - Count Descendants