#include<bits/stdc++.h>
#include<vector>
using namespace std;
 
// Returna total number of palindrome substring of
// length greater then equal to 2
int CountPS(vector<char> str, int n)
{
    // creat empty 2-D matrix that counts all palindrome
    // substring. dp[i][j] stores counts of palindromic
    // substrings in st[i..j]
    int dp[n][n];
    memset(dp, 0, sizeof(dp));
 
    // P[i][j] = true if substring str[i..j] is palindrome,
    // else false
    bool P[n][n];
    memset(P, false , sizeof(P));
 
    // palindrome of single lenght
    for (int i= 0; i< n; i++)
        P[i][i] = true;
 
    // palindrome of length 2
    for (int i=0; i<n-1; i++)
    {
        if (str[i] == str[i+1])
        {
            P[i][i+1] = true;
            dp[i][i+1] = 1 ;
        }
    }
 
    // Palindromes of length more then 2. This loop is similar
    // to Matrix Chain Multiplication. We start with a gap of
    // length 2 and fill DP table in a way that gap between
    // starting and ending indexes increases one by one by
    // outer loop.
    for (int gap=2 ; gap<n; gap++)
    {
        // Pick starting point for current gap
        for (int i=0; i<n-gap; i++)
        {
            // Set ending point
            int j = gap + i;
 
            // If current string is palindrome
            if (str[i] == str[j] && P[i+1][j-1] )
                P[i][j] = true;
 
            // Add current palindrome substring ( + 1)
            // and rest palinrome substring (dp[i][j-1] + dp[i+1][j])
            // remove common palinrome substrings (- dp[i+1][j-1])
            if (P[i][j] == true)
                dp[i][j] = dp[i][j-1] + dp[i+1][j] + 1 - dp[i+1][j-1];
            else
                dp[i][j] = dp[i][j-1] + dp[i+1][j] - dp[i+1][j-1];
        }
    }
 
    // return total palindromic substrings
    return dp[0][n-1];
}
 
// Driver program
int main()
{
    string s;
    cin>>s;
    int q,l,r;
    cin>>q;
    while(q--){
        vector<char>f;
        cin>>l>>r;
        for(int i=l-1;i<r;i++){
            f.push_back(s[i]);
        }
        cout << CountPS(f,f.size()) << endl;
        
    }
    return 0;
}