Problem: Ilya the Lion wants to help all his friends with passing exams. They need to solve the following problem to pass the IT exam.

You've got string s = s1s2... sn (n is the length of the string), consisting only of characters "." and "#" and m queries. Each query is described by a pair of integers li, ri (1 ≤ li < ri ≤ n). The answer to the query li, ri is the number of such integers i (li ≤ i < ri), that si = si + 1.

Ilya the Lion wants to help his friends but is there anyone to help him? Help Ilya, solve the problem.

Input

The first line contains string s of length n (2 ≤ n ≤ 105). It is guaranteed that the given string only consists of characters "." and "#".

The next line contains integer m (1 ≤ m ≤ 105) — the number of queries. Each of the next m lines contains the description of the corresponding query. The i-th line contains integers li, ri (1 ≤ li < ri ≤ n).

Output

Print m integers — the answers to the queries in the order in which they are given in the input.

Examples

Input

......
4
3 4
2 3
1 6
2 6


Output

1
1
5
4


Input

#..###
5
1 3
5 6
1 5
3 6
3 4


Output

1
1
2
2
0

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

public class Smile {
	static Scanner in;
	static PrintWriter out;
	static int[] c;
	
	static void func(int l, int r) {
		out.println(c[r - 1] - c[l - 1]);
		return;
	}
	
	public static void main(String[] args) throws IOException {
		in = new Scanner(System.in);
		out = new PrintWriter(System.out);
//		Map<String, Integer> hm = new HashMap<String, Integer>();
		String a = in.nextLine();
		int s = a.length();
		int[] b = new int[s];
		for (int i = 0; i < s - 1; i++) {
			if (a.charAt(i) == a.charAt(i + 1)) {
				b[i] = 1;
			} else {
				b[i] = 0;
			}
		}
		out.println();
		c = new int[s];
//		int[] d = new int[s - 1];
		c[0] = 0;
		for (int i = 1; i < s; i++) {
			c[i] = c[i - 1] + b[i - 1];
		} 
		int n = in.nextInt();
		for (int i = 0; i < n; i++) {
			int x = in.nextInt();
			int y = in.nextInt();
			func(x, y);
		}
		out.close();
		return;
	}
}