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:
/******************************************************************************

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Write your code in this editor and press "Run" button to execute it.

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import java.io.*;
import java.util.*;
public class Main
{
	public static void main(String[] args) throws Exception{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String s = br.readLine().trim();
		int[] prefix = new int[s.length()];
		prefix[0] = 0;
		for(int i = 1; i < prefix.length; i++){
	        if(s.charAt(i)==(s.charAt(i-1))){
	            prefix[i] = prefix[i-1]+1;
	        }
	        else{
	            prefix[i] = prefix[i-1];
	        }
	    }
		int m = Integer.parseInt(br.readLine().trim());
		for(int i = 0; i < m; i++){
		    String[] str = br.readLine().trim().split(" ");
		    int a = Integer.parseInt(str[0]);
		    int b = Integer.parseInt(str[1]);
		    System.out.println(prefix[b-1]-prefix[a-1]);
		}
	}
}