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:
/**
 *
 * @author Faruk
 */

import java.util.Arrays;
import java.util.Scanner;
import java.util.HashSet;
import java.util.HashMap;
import java.util.ArrayList;


public class tmp {
    public static void main(String [] args){
        Scanner scan = new Scanner(System.in);
        
        String n = scan.next();
        int [] arr = new int[n.length()+1];
        arr[0] = 0;
        int m = scan.nextInt();
        int l = 0;
        int r = 0;

        for(int i=1; i<n.length(); i++){
            if(n.charAt(i-1) == n.charAt(i))
                arr[i] = arr[i-1] + 1;
            else
                arr[i] = arr[i-1];
        }  

        StringBuilder out = new StringBuilder();
        for(int i=0; i<m; i++){
            l = scan.nextInt();
            r = scan.nextInt();
            out.append(arr[r-1]-arr[l-1]).append("\n");
        }
        System.out.println(out.toString());
                
                
                
                
                


        
        

        
        
        
        






        
        
        
        
    }

    
    
    
    
}