Problem: Mishka's favourite experimental indie band has recently dropped a new album! Songs of that album share one gimmick. Each name s_i is one of the following types:

  * 1~c — a single lowercase Latin letter; 
  * 2~j~c — name s_j (1 ≤ j < i) with a single lowercase Latin letter appended to its end. 



Songs are numbered from 1 to n. It's guaranteed that the first song is always of type 1.

Vova is rather interested in the new album but he really doesn't have the time to listen to it entirely. Thus he asks Mishka some questions about it to determine if some song is worth listening to. Questions have the following format:

  * i~t — count the number of occurrences of string t in s_i (the name of the i-th song of the album) as a continuous substring, t consists only of lowercase Latin letters. 



Mishka doesn't question the purpose of that information, yet he struggles to provide it. Can you please help Mishka answer all Vova's questions?

Input

The first line contains a single integer n (1 ≤ n ≤ 4 ⋅ 10^5) — the number of songs in the album.

Each of the next n lines contains the desciption of the i-th song of the album in the following format:

  * 1~c — s_i is a single lowercase Latin letter; 
  * 2~j~c — s_i is the name s_j (1 ≤ j < i) with a single lowercase Latin letter appended to its end. 



The next line contains a single integer m (1 ≤ m ≤ 4 ⋅ 10^5) — the number of Vova's questions.

Each of the next m lines contains the desciption of the j-th Vova's question in the following format:

  * i~t (1 ≤ i ≤ n, 1 ≤ |t| ≤ 4 ⋅ 10^5) — count the number of occurrences of string t in s_i (the name of the i-th song of the album) as a continuous substring, t consists only of lowercase Latin letters. 



It's guaranteed that the total length of question strings t doesn't exceed 4 ⋅ 10^5.

Output

For each question print a single integer — the number of occurrences of the question string t in the name of the i-th song of the album as a continuous substring.

Example

Input


20
1 d
2 1 a
2 2 d
2 3 a
2 4 d
2 5 a
2 6 d
2 7 a
1 d
2 9 o
2 10 k
2 11 i
2 12 d
2 13 o
2 14 k
2 15 i
2 1 o
2 17 k
2 18 i
2 15 i
12
8 da
8 dada
8 ada
6 dada
3 dada
19 doki
19 ok
16 doki
15 doki
9 d
1 a
20 doki


Output


4
3
3
2
0
1
1
2
1
1
0
2

Note

Song names of the first example:

  1. d 
  2. da 
  3. dad 
  4. dada 
  5. dadad 
  6. dadada 
  7. dadadad 
  8. dadadada 
  9. d 
  10. do 
  11. dok 
  12. doki 
  13. dokid 
  14. dokido 
  15. dokidok 
  16. dokidoki 
  17. do 
  18. dok 
  19. doki 
  20. dokidoki 



Thus the occurrences for each question string are:

  1. string "da" starts in positions [1, 3, 5, 7] in the name "dadadada"; 
  2. string "dada" starts in positions [1, 3, 5] in the name "dadadada"; 
  3. string "ada" starts in positions [2, 4, 6] in the name "dadadada"; 
  4. string "dada" starts in positions [1, 3] in the name "dadada"; 
  5. no occurrences of string "dada" in the name "dad"; 
  6. string "doki" starts in position [1] in the name "doki"; 
  7. string "ok" starts in position [2] in the name "doki"; 
  8. string "doki" starts in positions [1, 5] in the name "dokidoki"; 
  9. string "doki" starts in position [1] in the name "dokidok"; 
  10. string "d" starts in position [1] in the name "d"; 
  11. no occurrences of string "a" in the name "d"; 
  12. string "doki" starts in positions [1, 5] in the name "dokidoki". 

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.StringTokenizer;

public class AhoOnTrie {
	
	static BIT treeState;

	public static void main(String[] args) {
		FastScanner fs=new FastScanner();
		int nSongs=fs.nextInt();
		TrieNode[] songs=new TrieNode[nSongs+1];
		songs[0]=new TrieNode();
		for (int i=1; i<=nSongs; i++) {
			int type=fs.nextInt();
			int parent=type==1?0:fs.nextInt();
			char added=fs.next().charAt(0);
			TrieNode newNode=songs[parent].add(added);
			songs[i]=newNode;
		}
		int nQueries=fs.nextInt();
		String[] queryStrings=new String[nQueries];
		Query[] queries=new Query[nQueries];
		for (int qq=0; qq<nQueries; qq++) {
			int node=fs.nextInt();
			String queryString=fs.next();
			queryStrings[qq]=queryString;
			songs[node].queriesAtNode.add(queries[qq]=new Query(qq, queryString));
		}
		AhoCorasick.createAhoCorasick(queryStrings);
		
		//make sLink tree
		Node[] sLinkTree=new Node[AhoCorasick.nodeCount];
		for (int i=0; i<sLinkTree.length; i++) sLinkTree[i]=new Node();
		for (int i=1; i<sLinkTree.length; i++) {
			int par=AhoCorasick.suffixLink[i];
			sLinkTree[par].children.add(sLinkTree[i]);
		}
		sLinkTree[0].dfs(-1);
		treeState=new BIT(sLinkTree[0].postorderNumber+1);
		
		songs[0].answerDFS(0, sLinkTree);
		
		PrintWriter out=new PrintWriter(System.out);
		for (int i=0; i<nQueries; i++)
			out.println(queries[i].ans);
		out.close();
	}
	
	static class Query {
		int queryNumber;
		String queryWord;
		int ans;
		public Query(int queryNumber, String queryWord) {
			this.queryNumber=queryNumber;
			this.queryWord=queryWord;
		}
	}
	
	//trie given as input
	static class TrieNode {
		TrieNode[] adj=new TrieNode[26];
		ArrayList<Query> queriesAtNode=new ArrayList<>();
		
		TrieNode add(char c) {
			int ind=c-'a';
			if (adj[ind]==null) adj[ind]=new TrieNode();
			return adj[ind];
		}
		
		void answerDFS(int ahoIndex, Node[] sLinkTree) {
			//increment the number of times this node has been visited 
			Node sLinkNode=sLinkTree[ahoIndex];
			treeState.rangeUpdate(0, sLinkNode.preorderNumber, 1);
			
			//answer all queries at this node
			for (Query q:queriesAtNode) {
				//TODO: answer
				int curAhoNode=0;
				for (char c:q.queryWord.toCharArray())
					curAhoNode=AhoCorasick.transition[c-'a'][curAhoNode];
				Node querySLinkNode=sLinkTree[curAhoNode];
				int ans=treeState.pointQuery(querySLinkNode.preorderNumber)-treeState.pointQuery(querySLinkNode.postorderNumber);
				q.ans=ans;
			}
			
			//go to all children, changing the aho index
			for (int kid=0; kid<26; kid++) {
				if (adj[kid]==null) continue;
				adj[kid].answerDFS(AhoCorasick.transition[kid][ahoIndex], sLinkTree);
			}
			
			//decrement the number of times this node has been visited
			treeState.rangeUpdate(0, sLinkNode.preorderNumber, -1);
		}
	}
	
	//tree node for aho s-links
	static class Node {
		ArrayList<Node> children=new ArrayList<>();
		int preorderNumber, postorderNumber;
		public int dfs(int preNumber) {
			this.preorderNumber=++preNumber;
			for (Node n:children) {
				preNumber=n.dfs(preNumber);
			}
			this.postorderNumber=++preNumber;
			return preNumber;
		}
	}
	
	//builds aho for query strings
	static class AhoCorasick {
		static final int ALPHA = 26;
		static int nodeCount, transition[][];
		static int[] suffixLink;
		static boolean[] term;
		static void createAhoCorasick(String[] strs) {
			int maxNodes = 1;
			for(String s : strs) maxNodes += s.length();

			int[][] children = new int[ALPHA][maxNodes];
			boolean[] leaf = new boolean[maxNodes];
			nodeCount = 1;
			for(String s : strs) {
				int node = 0;
				for(char ch : s.toCharArray()) {
					int c = ch - 'a';
					if(children[c][node] == 0) children[c][node] = nodeCount++;
					node = children[c][node];
				}
				leaf[node] = true;
			}

			transition = new int[ALPHA][nodeCount];
			term = new boolean[nodeCount];
			suffixLink=new int[nodeCount];

			ArrayDeque<Integer> queue = new ArrayDeque<>();
			queue.add(0);  queue.add(0);

			while(queue.size() > 0) {
				int node = queue.remove(), suffLink = queue.remove();
				suffixLink[node]=suffLink;

				term[node] = leaf[node] || term[suffLink];

				for(int ch = 0; ch < ALPHA; ch++) {
					if(children[ch][node] != 0){
						transition[ch][node] = children[ch][node];
						queue.add(children[ch][node]);
						queue.add(node == 0 ? 0 : transition[ch][suffLink]);
					} else  transition[ch][node] = transition[ch][suffLink];
				}
			}
		}
	}
	
	//for tree updates
	static class BIT {
		int n;
		int[] tree;
		
		public BIT(int n) {
			this.n = n;
			tree = new int[n + 5];
		}
		
		int read(int i) {
			i++;
			int sum = 0;
			while (i > 0) {
				sum += tree[i];
				i -= i & -i;
			}
			return sum;
		}
		
		void update(int i, int val) {
			i++;
			while (i <= n) {
				tree[i] += val;
				i += i & -i;
			}
		}
		
		void rangeUpdate(int from, int to, int delta) {
			update(from, delta);
			update(to+1, -delta);
		}
		
		int pointQuery(int i) {
			return read(i);
		}

		// if the BIT is a freq array, returns the
		// index of the kth item, or n if there are fewer
		// than k items.
		int getKth(int k) {
			int e=Integer.highestOneBit(n), o=0;
			for (; e!=0; e>>=1) {
				if (e+o<=n && tree[e+o]<=k) {
					k-=tree[e+o];
					o+=e;
				}
			}
			return o;
		}
		
	}


	
	static class FastScanner {
		StringTokenizer st=new StringTokenizer("");
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		String next() {
			while (!st.hasMoreElements())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		public int nextInt() {
			return Integer.parseInt(next());
		}
	}
}