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 H {

	String s;
	int[][] f;
		
    public void solve() throws IOException {
        s = nextToken();
        init();
        int q = nextInt();
        for (int i = 0; i < q; i++) {
        	int l = nextInt() - 1;
        	int r = nextInt() - 1;
        	System.out.println(f[l][r] / 2);
        }
    }
    
    int n;
    void init() {
    	n = s.length();
    	f = new int[n][n];
    	for (int i = 0; i < n; i++) 
    		f[i][i] = 1 * 2 + 1;
    	for (int i = 0; i < n - 1; i++)
    		if (s.charAt(i) == s.charAt(i + 1)) 
    			f[i][i + 1] = 3 * 2 + 1;
    		else f[i][i + 1] = 2 * 2 + 0;
    	//System.out.println(f[1][1]);
    	for (int l = 2; l < n; l++)
    		for (int i = 0; i + l < n; i++) {
    			int j = i + l;
    			int v1 = f[i + 1][j] / 2;
    			int v2 = f[i][j - 1] / 2;
    			int v3 = f[i + 1][j - 1] / 2;
    			f[i][j] = (v1 + v2 - v3) * 2;
    			if (s.charAt(i) == s.charAt(j) && f[i + 1][j - 1] % 2 == 1) {
    				f[i][j] = (f[i][j] / 2 + 1) * 2 + 1;
    			}
    			//System.out.println(i + " " + j + " " + f[i][j] / 2);
    		}
    }
    
    public static void main(String[] args) throws FileNotFoundException {
        new H().run();
    }

    BufferedReader reader;
    StringTokenizer tokenizer;
    PrintWriter writer;

    public void run() {
        try {
            long tbegin = System.currentTimeMillis();
            reader = new BufferedReader(new InputStreamReader(System.in));
            //reader = new BufferedReader(new InputStreamReader(new FileInputStream("art.in")));
            tokenizer = null;
            writer = new PrintWriter(System.out);
            //writer = new PrintWriter(new FileOutputStream("art.out"));
            solve();
            //reader.close();
            //System.out.println(System.currentTimeMillis() - tbegin + "ms");
            writer.close();
        } catch (Exception e) {
            e.printStackTrace();
            System.exit(1);
        }
    }

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

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

    double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }

    String nextToken() throws IOException {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            tokenizer = new StringTokenizer(reader.readLine());
        }
        return tokenizer.nextToken();
    }

}