Problem: Polycarpus got hold of a family tree. The found tree describes the family relations of n people, numbered from 1 to n. Every person in this tree has at most one direct ancestor. Also, each person in the tree has a name, the names are not necessarily unique.

We call the man with a number a a 1-ancestor of the man with a number b, if the man with a number a is a direct ancestor of the man with a number b.

We call the man with a number a a k-ancestor (k > 1) of the man with a number b, if the man with a number b has a 1-ancestor, and the man with a number a is a (k - 1)-ancestor of the 1-ancestor of the man with a number b.

In the tree the family ties do not form cycles. In other words there isn't a person who is his own direct or indirect ancestor (that is, who is an x-ancestor of himself, for some x, x > 0).

We call a man with a number a the k-son of the man with a number b, if the man with a number b is a k-ancestor of the man with a number a.

Polycarpus is very much interested in how many sons and which sons each person has. He took a piece of paper and wrote m pairs of numbers vi, ki. Help him to learn for each pair vi, ki the number of distinct names among all names of the ki-sons of the man with number vi.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 105) — the number of people in the tree. Next n lines contain the description of people in the tree. The i-th line contains space-separated string si and integer ri (0 ≤ ri ≤ n), where si is the name of the man with a number i, and ri is either the number of the direct ancestor of the man with a number i or 0, if the man with a number i has no direct ancestor. 

The next line contains a single integer m (1 ≤ m ≤ 105) — the number of Polycarpus's records. Next m lines contain space-separated pairs of integers. The i-th line contains integers vi, ki (1 ≤ vi, ki ≤ n).

It is guaranteed that the family relationships do not form cycles. The names of all people are non-empty strings, consisting of no more than 20 lowercase English letters.

Output

Print m whitespace-separated integers — the answers to Polycarpus's records. Print the answers to the records in the order, in which the records occur in the input.

Examples

Input

6
pasha 0
gerald 1
gerald 1
valera 2
igor 3
olesya 1
5
1 1
1 2
1 3
3 1
6 1


Output

2
2
0
1
0


Input

6
valera 0
valera 1
valera 1
gerald 0
valera 4
kolya 4
7
1 1
1 2
2 1
2 2
4 1
5 1
6 1


Output

1
0
0
0
2
0
0

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

public class b {
	
	static ArrayList<pair>[] q;
	static ArrayList<Integer>[] nxt;
	static int[]ans, size, cnt;
	static int size(int u) {
		int c = 1;
		for(int v : nxt[u])
			c += size(v);
		return size[u] = c;
	}
	static void dfs( int u, int h, boolean f) {
		int w = -1; int max = -1;
		for(int v : nxt[u])
			if(size[v] > max) {
				max = size[v];
				w = v;
			}
		for(int v : nxt[u])
			if(v != w)
				dfs(v,h+1,false);
		if(w != -1)
			dfs(w,h+1,true);
		add(u,h,w);
		for(pair p : q[u]) {
			ans[p.u] = cnt[h+p.k] - 1;
		}
		if(!f)
			erase(u,h);
	}
	static void add(int u,int h,int w) {
		cnt[h]++;
		for(int v : nxt[u])
			if(v != w)
				add(v,h+1,w);
	}
	static void erase(int u,int h) {
		cnt[h]--;
		for(int v : nxt[u])
			erase(v,h+1);
	}
	
	public static void main(String[] args) throws IOException, InterruptedException {
		Scanner zizo = new Scanner(System.in);
		PrintWriter wr = new PrintWriter(System.out);
		
		N = n = zizo.nextInt();
		dp = new int[18][N];
		for(int i = 0;i < n; i++)
			dp[0][i] = zizo.nextInt()-1;
		preprocess();
		int m = zizo.nextInt();
		q = new ArrayList[N];
		nxt = new ArrayList[N];
		for(int i = 0;i < N; i++) {
			q[i] = new ArrayList<>();
			nxt[i] = new ArrayList<>();
		}
		for(int i = 0;i < N; i++)
			if(dp[0][i] != -1)
				nxt[dp[0][i]].add(i);
		for(int i = 0;i < m; i++) {
			int v = zizo.nextInt()-1;
			int p = zizo.nextInt();
			v = getKth(p,v);
			if(v != -1)
				q[v].add(new pair(i,p));
		}
		ans = new int[m]; size = new int[N]; cnt = new int[N];
		for(int i = 0;i < N; i++)
			if(dp[0][i] == -1) {
				size(i);
				dfs(i,0,false);
			}
		for(int x : ans)
			wr.print(x+" ");
		
		wr.close();
	}
	static int getKth(int k, int u)
	{
		for(int i = 17; i >= 0; --i)
			if(1<<i <= k && dp[i][u] != -1)
			{
				u = dp[i][u];
				k ^= 1<<i;
			}
		return k == 0 ? u : -1;
	}
	//zizo's algorithm of LCA
	static int N;
	static int l; //log(N)
	static int n,dpth[],dp[][]; //dpth[N] && dp[l + 1][N]
	static ArrayList<Integer>[]adj;
	static void dfs1(int u,int p) {
		dp[0][u] = p;	//dp[0][root] = -1
		
		for(int v : adj[u])
			if(v != p) {
				dpth[v] = dpth[u] + 1;	//dpth[root] = 0
				dfs1(v , u);
			}
	}
	
	static int getKthAncestor(int u, int k) {
		int d = dpth[u] - k;	//dpth of kth ancestor
		
		for(int j = l;j >= 0; j--) {
			int nu = dp[j][u];
			if(nu == -1)
				continue;
			if(dpth[nu] == d)
				return nu;
			else if(dpth[nu] > d)
				u = nu;
		}
		return -1;
	}
	
	static int getLCA(int a, int b) {
		if(dpth[a] < dpth[b]) { a ^= b; b ^= a; a ^= b; }
		a = getKthAncestor(a, dpth[a] - dpth[b]);
		
		//dpth[a] == dpth[b]
		if(a == b)
			return a;
		for(int j = l;j >= 0; j--) {
			int na = dp[j][a], nb = dp[j][b];
			if(na != -1 && nb != -1 && na != nb) {
				a = na; 
				b = nb;
			}
		}
		return dp[0][a];
	}
	static void preprocess(){
		l = 17;
		//dp = new int[l + 1][N];
		dpth = new int[N];
//		for(int[] a : dp)
//			Arrays.fill(a, -1);
		//dfs(0,-1);
		for(int j = 1; j <= l; j++)
			for(int i = 0;i < n; i++)
				dp[j][i] = (dp[j-1][i] == -1 ? -1 : dp[j - 1][dp[j - 1][i]]);
	}
}
class pair{
	int u,k;
	pair(int a,int b){u=a;k=b;}
}

class Scanner 
{
	StringTokenizer st;
	BufferedReader br;

	public Scanner(InputStream s){	br = new BufferedReader(new InputStreamReader(s));}

	public String next() throws IOException 
	{
		while (st == null || !st.hasMoreTokens()) 
			st = new StringTokenizer(br.readLine());
		return st.nextToken();
	}

	public int nextInt() throws IOException {return Integer.parseInt(next());}

	public long nextLong() throws IOException {return Long.parseLong(next());}

	public String nextLine() throws IOException {return br.readLine();}

	public double nextDouble() throws IOException
	{
		String x = next();
		StringBuilder sb = new StringBuilder("0");
		double res = 0, f = 1;
		boolean dec = false, neg = false;
		int start = 0;
		if(x.charAt(0) == '-')
		{
			neg = true;
			start++;
		}
		for(int i = start; i < x.length(); i++)
			if(x.charAt(i) == '.')
			{
				res = Long.parseLong(sb.toString());
				sb = new StringBuilder("0");
				dec = true;
			}
			else
			{
				sb.append(x.charAt(i));
				if(dec)
					f *= 10;
			}
		res += Long.parseLong(sb.toString()) / f;
		return res * (neg?-1:1);
	}

	public boolean ready() throws IOException {return br.ready();}
}