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».
// practice with kaiboy
import java.io.*;
import java.util.*;

public class CF245H extends PrintWriter {
	CF245H() { super(System.out); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF245H o = new CF245H(); o.main(); o.flush();
	}

	void manacher(int[] rr, byte[] cc, int n) {
		for (int o = -1, x = 0, i = 0; i < n; i++)
			if (o + o - i >= 0 && i + rr[o + o - i] < x)
				rr[i] = rr[o + o - i];
			else {
				o = i;
				if (x < o + 1)
					x = o + 1;
				while (x < n && o + o - x >= 0 && cc[x] == cc[o + o - x])
					x++;
				rr[i] = x - o;
			}
	}
	void main() {
		byte[] aa = sc.next().getBytes();
		int n = aa.length;
		int m = n * 2 + 1;
		byte[] bb = new byte[m];
		for (int i = 0; i < n; i++)
			bb[i * 2 + 1] = aa[i];
		int[] rr = new int[m];
		manacher(rr, bb, m);
		int[][] dp = new int[n][n];
		for (int h = 0; h < m; h++) {
			int i = h - rr[h] + 1;
			int j = h + rr[h] - 1;
			for (i = i / 2, j = j / 2 - 1; i <= j; i++, j--)
				dp[i][j] = 1;
		}
		for (int i = n - 1; i >= 0; i--)
			for (int j = i + 1; j < n; j++)
				dp[i][j] += dp[i + 1][j] + dp[i][j - 1] - dp[i + 1][j - 1];
		int q = sc.nextInt();
		while (q-- > 0) {
			int l = sc.nextInt() - 1;
			int r = sc.nextInt() - 1;
			println(dp[l][r]);
		}
	}
}
