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.util.*;
import java.io.*;
public class Main {
public static void main(String [] args)throws Exception{
	BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	String s = br.readLine();
	int q = Integer.parseInt(br.readLine());
	int dp[][] = new int[s.length() + 2][s.length() + 2];
	boolean isPal[][] = new boolean[s.length() + 2][s.length() + 2];
	for(int i = s.length();i >= 1;i--){
		for(int j = s.length();j >= i;j--){
			if(s.charAt(i - 1) == s.charAt(j - 1) && (isPal[i + 1][j - 1] || i + 1 >= j - 1))
				                                      isPal[i][j] = true;
		}
	}
	for(int i = s.length();i >= 1;i--){
		for(int j = i;j <= s.length();j++){
			dp[i][j] = dp[i + 1][j] + dp[i][j - 1] - dp[i + 1][j - 1];
			if(s.charAt(i - 1) == s.charAt(j - 1) && (isPal[i + 1][j - 1] 
				                                            || i + 1 >= j - 1))dp[i][j]++;
		}
	}
	StringTokenizer st;
	StringBuilder sb = new StringBuilder();
	for(int i = 0;i < q;i++){
		st = new StringTokenizer(br.readLine());
		int l = Integer.parseInt(st.nextToken());
		int r = Integer.parseInt(st.nextToken());
		sb.append(dp[l][r]);
		sb.append('\n');
	}
	System.out.print(sb);
}
}