#include <bits/stdc++.h>
using namespace std;


void palindromeSubStrs(string s)
{
    map<string, int> m;
    int n = s.size();
 
    // table for storing results (2 rows for odd-
    // and even-length palindromes
    int R[2][n+1];
 
    // Find all sub-string palindromes from the given input
    // string insert 'guards' to iterate easily over s
    s = "@" + s + "#";
 
    for (int j = 0; j <= 1; j++)
    {
        int rp = 0;   // length of 'palindrome radius'
        R[j][0] = 0;
 
        int i = 1;
        while (i <= n)
        {
            //  Attempt to expand palindrome centered at i
            while (s[i - rp - 1] == s[i + j + rp])
                rp++;  // Incrementing the length of palindromic
                       // radius as and when we find vaid palindrome
 
            // Assigning the found palindromic length to odd/even
            // length array
            R[j][i] = rp;
            int k = 1;
            while ((R[j][i - k] != rp - k) && (k < rp))
            {
                R[j][i + k] = min(R[j][i - k],rp - k);
                k++;
            }
            rp = max(rp - k,0);
            i += k;
        }
    }
 
    // remove 'guards'
    s = s.substr(1, n);
 
    // Put all obtained palindromes in a hash map to
    // find only distinct palindromess
    m[string(1, s[0])]=1;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 0; j <= 1; j++)
            for (int rp = R[j][i]; rp > 0; rp--)
               m[s.substr(i - rp - 1, 2 * rp + j)]=1;
        m[string(1, s[i])]=1;
    }
 
    //printing all distinct palindromes from hash map
   //cout << "Below are " << m.size()-1
     //   << " palindrome sub-strings";
   map<string, int>::iterator ii;
    map<int,int>ab;
   for (ii = m.begin(); ii!=m.end(); ++ii)
   {
        string sx=(*ii).first;
       int len=sx.length();
       ab[len]++;
   } 
    map<int,int>::reverse_iterator rit;
    rit=ab.rbegin();
    cout<<rit->second<<"\n";
}

/*
string findLongestPalindrome(string str)
{
    // to stores freq of characters in a string
    int count[256] = { 0 };
 
    // find freq of characters in the input string
    for (int i = 0; i < str.size(); i++)
        count[str[i]]++;
 
    // Any palindromic string consists of three parts
    // beg + mid + end
    string beg = "", mid = "", end = "";
 
    // solution assumes only lowercase characters are
    // present in string. We can easily extend this
    // to consider any set of characters
    for (char ch = 'a'; ch <= 'z'; ch++)
    {
        // if the current character freq is odd
        if (count[ch] & 1)
        {
            // mid will contain only 1 character. It
            // will be overridden with next character
            // with odd freq
            mid = ch;
 
            // decrement the character freq to make
            // it even and consider current character
            // again
            count[ch--]--;
        }
 
        // if the current character freq is even
        else
        {
            // If count is n(an even number), push
            // n/2 characters to beg string and rest
            // n/2 characters will form part of end
            // string
            for (int i = 0; i < count[ch]/2 ; i++)
                beg.push_back(ch);
        }
    }
 
    // end will be reverse of beg
    end = beg;
    reverse(end.begin(), end.end());
 
    // return palindrome string
    beg + mid + end;
} */

int main() {
    string str;
    cin>>str;
    int T;
    cin>>T;
    while(T--)
    {
        int l,r;
        cin>>l>>r;
        string str1="";
        for(int i=l-1;i<r;i++)
            str1+=str[i];
         palindromeSubStrs(str1);
    }
    
    return 0;
}