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.*;
import java.util.*;

public class H {
    public static void main(String[] args) {
        try {
            new H().solve();
        } catch (Exception e) {
            e.printStackTrace();
        }
    }

    void solve() throws IOException {
        // Scanner sc = new Scanner(System.in);
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        String[] sp;

        String str = in.readLine();
        int n = str.length();
        int[][] dp = new int[n][n];
        boolean[][] pal = new boolean[n][n];
        for (int i = 0; i < n; i++) {
            pal[i][i] = true;
        }
        for (int i = 0; i < n - 1; i++) {
            pal[i][i + 1] = str.charAt(i) == str.charAt(i + 1);
        }
        for (int w = 2; w < n; w++) {
            for (int i = 0; i + w < n; i++) {
                pal[i][i + w] =
                        pal[i + 1][i + w - 1] &&
                                str.charAt(i) == str.charAt(i + w);
            }
        }

        for (int i = 0; i < n; i++) {// 0,1
            dp[i][i] = 1;
            if (i + 1 < n) {
                dp[i][i + 1] = 2 + (pal[i][i + 1] ? 1 : 0);
            }
        }
        for (int w = 2; w < n; w++) {
            for (int i = 0; i + w < n; i++) {
                int a = i;
                int b = i + w;
                dp[a][b] = dp[a + 1][b]
                        + (dp[a][b - 1] - dp[a + 1][b - 1] + (pal[a][b] ? 1 : 0));
            }
        }

        str = null;
        pal = null;

        int Q = Integer.parseInt(in.readLine());
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < Q; i++) {
            sp = in.readLine().split(" ");
            int l = Integer.parseInt(sp[0]) - 1;
            int r = Integer.parseInt(sp[1]) - 1;
            // System.out.println(dp[l][r]);
            sb.append(dp[l][r] + "\n");
        }
        System.out.print(sb);
    }
}
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