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.*;
import java.math.*;
import static java.lang.Math.*;
import static java.lang.Integer.parseInt;
import static java.lang.Long.parseLong;
import static java.lang.Double.parseDouble;
import static java.lang.String.*;

public class Main {
    
    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        StringBuilder out = new StringBuilder();
        StringTokenizer tk;
        
        String s = in.readLine();
        int n = s.length();
        
        boolean [][] isPal = new boolean[n][n];
        int [][] dp = new int[n][n];
        
        for(int i=0; i<n-1; i++) {
            isPal[i][i] = true;
            dp[i][i] = 1;
            if(s.charAt(i)==s.charAt(i+1)) {
                isPal[i][i+1] = true;
                dp[i][i+1] = 3;
            } else dp[i][i+1] = 2;
        }
        isPal[n-1][n-1] = true;
        dp[n-1][n-1] = 1;
        
        for(int len=3; len<=n; len++) {
            for(int i=0; i<=n-len; i++) {
                int j = i+len-1;
                isPal[i][j] = isPal[i+1][j-1] && s.charAt(i)==s.charAt(j);
            }
        }
        
        
        for(int len=3; len<=n; len++) {
            for(int i=0; i<=n-len; i++) {
                int j = i+len-1;
                dp[i][j] = dp[i+1][j]+dp[i][j-1]-dp[i+1][j-1]+(isPal[i][j] ? 1 : 0);
            }
        }
        
        int q = parseInt(in.readLine());
        
        while(q-- > 0) {
            tk = new StringTokenizer(in.readLine());
            int l = parseInt(tk.nextToken())-1,r = parseInt(tk.nextToken())-1;
            
            out.append(dp[l][r]).append("\n");
        }
        
        System.out.print(out);
    }
}
