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 problems {
    public static void main(String args[] ) throws Exception {
        FastReader sc=new FastReader();
        String s=sc.nextLine();
        int n=s.length();
        s=" "+s;
        int[][] dp=new int[5001][5001];

        // Length of 1
        for(int i=1;i<=n;i++){
            dp[i][i]=1;
        }

        // Length of 2
        for(int i=1;i<n;i++){
            if(s.charAt(i)== s.charAt(i+1)){
                dp[i][i+1]=1;
            }
        }

        for(int len=0;len<n-2;len++){
            //i is a starting index
            for(int i=1;i<n-len;i++){
                //if we have a palindrome ans the same letter on both sides of the palindrome
                if(dp[i][i+len]==1 && s.charAt(i-1)== s.charAt(i+len+1)){
                    dp[i-1][i+len+1]=1;
                }
            }
        }

        // Prefix Sum 2-d
        for (int i=1; i <= n; i++) {
            for (int j = 2; j <= n; j++) {
                dp[i][j]=dp[i][j]+dp[i][j-1];
            }
        }
        for (int i=1; i <= n; i++) {
            for (int j = 2; j <= n; j++) {
                dp[j][i]=dp[j][i]+dp[j-1][i];
            }
        }
        StringBuilder sb=new StringBuilder();
        int q=sc.nextInt();
        while(q-->0){
            int x=sc.nextInt();
            int y=sc.nextInt();
            int ans=0;

            ans=dp[y][y] - dp[x-1][y] - dp[y][x-1] + dp[x-1][x-1];
            sb.append(ans+"\n");
        }
        System.out.println(sb);
    }
}
class FastReader
{
    BufferedReader br;
    StringTokenizer st;

    public FastReader()
    {
        br = new BufferedReader(new
                InputStreamReader(System.in));
    }

    String next()
    {
        while (st == null || !st.hasMoreElements())
        {
            try
            {
                st = new StringTokenizer(br.readLine());
            }
            catch (IOException  e)
            {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }

    int nextInt()
    {
        return Integer.parseInt(next());
    }

    long nextLong()
    {
        return Long.parseLong(next());
    }

    double nextDouble()
    {
        return Double.parseDouble(next());
    }

    String nextLine()
    {
        String str = "";
        try
        {
            str = br.readLine();
        }
        catch (IOException e)
        {
            e.printStackTrace();
        }
        return str;
    }
}