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.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.StringTokenizer;


public class B {
    
    static StringTokenizer st;
    static BufferedReader in;
    public static void main(String[] args) throws IOException {
        in = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        String s = next();
        int n = s.length();
        int[][]dp = new int[n][n];
        boolean[][]f_odd = new boolean[n][n], f_even = new boolean[n][n];
        for (int i = 0; i < n; i++) {
            dp[i][i] = 1;
            f_odd[i][i] = true;
        }
        for (int i = 2; i <= n; i++) {
            for (int j = 0; j <= n-i; j++) {
                dp[j][j+i-1] = dp[j+1][j+i-1]+dp[j][j+i-2]-dp[j+1][j+i-2];
                if (s.charAt(j)==s.charAt(j+i-1)) {
                    if (i % 2==1) {
                        if (f_odd[j+1][j+i-2]) {
                            dp[j][j+i-1]++;
                            f_odd[j][j+i-1] = true;
                        }
                        else
                            f_odd[j][j+i-1] = false;
                    }
                    else {
                        if (i==2) {
                            dp[j][j+i-1]++;
                            f_even[j][j+i-1] = true;
                        }
                        else {
                            if (f_even[j+1][j+i-2]) {
                                dp[j][j+i-1]++;
                                f_even[j][j+i-1] = true;
                            }
                            else
                                f_even[j][j+i-1] = false;
                        }
                    }
                }
            }
        }
        int q = nextInt();
        for (int i = 1; i <= q; i++) {
            int L = nextInt();
            int R = nextInt();
            L--;
            R--;
            pw.println(dp[L][R]);
        }
        pw.close();
    }
    private static int nextInt() throws IOException{
        return Integer.parseInt(next());
    }
    
    private static long nextLong() throws IOException{
        return Long.parseLong(next());
    }
    
    private static double nextDouble() throws IOException{
        return Double.parseDouble(next());
    }
    
    private static String next() throws IOException {
        while (st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(in.readLine());
        }
        return st.nextToken();
    }
}
