You've got a string s = s1s2... s|s| of length |s|, consisting of lowercase English letters. There also are q queries, each query is described by two integers li, ri (1 ≤ li ≤ ri ≤ |s|). The answer to the query is the number of substrings of string s[li... ri], which are palindromes.

String s[l... r] = slsl + 1... sr (1 ≤ l ≤ r ≤ |s|) is a substring of string s = s1s2... s|s|.

String t is called a palindrome, if it reads the same from left to right and from right to left. Formally, if t = t1t2... t|t| = t|t|t|t| - 1... t1.

Input

The first line contains string s (1 ≤ |s| ≤ 5000). The second line contains a single integer q (1 ≤ q ≤ 106) — the number of queries. Next q lines contain the queries. The i-th of these lines contains two space-separated integers li, ri (1 ≤ li ≤ ri ≤ |s|) — the description of the i-th query.

It is guaranteed that the given string consists only of lowercase English letters.

Output

Print q integers — the answers to the queries. Print the answers in the order, in which the queries are given in the input. Separate the printed numbers by whitespaces.

Examples

Input

caaaba
5
1 1
1 4
2 3
4 6
4 5


Output

1
7
3
4
2

Note

Consider the fourth query in the first test case. String s[4... 6] = «aba». Its palindrome substrings are: «a», «b», «a», «aba».
import java.io.*;
import java.util.*;


public class C{

	public static void main(String[] args) throws IOException {
		Scanner sc=new Scanner();
		PrintWriter out=new PrintWriter(System.out);
		char []s=sc.next().toCharArray();
		int n=s.length;
		boolean [][]pal=new boolean[n][n];
		FenwickTree tree=new FenwickTree(n);
		ArrayList<int[]>[]queries=new ArrayList[n];
		for(int i=0;i<n;i++) {
			pal[i][i]=true;
			tree.update(i, 1);
			queries[i]=new ArrayList();
		}
		for(int len=2;len<=n;len++)
			for(int i=0;i+len<=n;i++) {
				int j=i+len-1;
				pal[i][j]=s[i]==s[j] && (len==2 || pal[i+1][j-1]);
				if(pal[i][j])
					tree.update(j, 1);
			}
		
		int q=sc.nextInt();
		for(int i=0;i<q;i++) {
			int l=sc.nextInt()-1,r=sc.nextInt()-1;
			queries[l].add(new int [] {r,i});
		}
		int []ans=new int [q];
		for(int l=0;l<n;l++) {
			for(int []query:queries[l]) {
				int r=query[0],idx=query[1];
				ans[idx]=tree.query(r);
			}
			for(int r=l;r<n;r++)
				if(pal[l][r])
					tree.update(r, -1);
		}
		for(int x:ans)
			out.println(x);
		out.close();

	}
	static class FenwickTree{
		int []bit;
		FenwickTree(int n){
			bit=new int [n+1];
		}
		void update(int idx,int v) {
			idx++;
			while(idx<bit.length) {
				bit[idx]+=v;
				idx+=(idx&-idx);
			}
		}
		int query(int idx) {
			idx++;
			int ans=0;
			while(idx>0) {
				ans+=bit[idx];
				idx-=(idx&-idx);
			}
			return ans;
		}
		
	}
	static class Scanner
	{
		BufferedReader br;
		StringTokenizer st;
		Scanner(){
			br=new BufferedReader(new InputStreamReader(System.in));
		}
		Scanner(String fileName) throws FileNotFoundException{
			br=new BufferedReader(new FileReader(fileName));
		}
		String next() throws IOException {
			while(st==null || !st.hasMoreTokens())
				st=new StringTokenizer(br.readLine());
			return st.nextToken();
		}
		String nextLine() throws IOException {
			return br.readLine();
		}
		int nextInt() throws IOException{
			return Integer.parseInt(next());
		}
		long nextLong()  throws NumberFormatException, IOException {
			return Long.parseLong(next());
		}
		double nextDouble() throws NumberFormatException, IOException {
			return Double.parseDouble(next());
		}
	}
}
