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.util.Scanner;
import java.util.Set;

public class Main {

    private static final int N = 100;
    private static final int M = 110;

    private static int m, n;
    private static int ax, bx, cx, dx;
    private static int cur, ans, obj;
    private static int[] da, ta;
    private static int[][] daa, taa;
    private static char[] ca;
    private static String as, bs, cs, ds;
    private static Set<Integer> set;
    private static Scanner in = new Scanner(System.in);

    public static void main(String[] args) {
        as = in.next();
        ca = as.toCharArray();
        n = ca.length;
        da = new int[n];


        for (int i = 1; i < n; i++) {
            da[i] = da[i-1] + (ca[i] == ca[i - 1] ? 1 : 0);
        }

        m = in.nextInt();
        for (int i = 0; i < m; i++) {
            ax = in.nextInt();
            bx = in.nextInt();
            System.out.println(da[bx-1]-da[ax-1]);
        }
    }

}