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* = *s*<sub class="lower-index">1</sub>*s*<sub class="lower-index">2</sub>... *s*<sub class="lower-index">*n*</sub> (*n* is the length of the string), consisting only of characters "." and "#" and *m* queries. Each query is described by a pair of integers *l*<sub class="lower-index">*i*</sub>, *r*<sub class="lower-index">*i*</sub> (1 ≤ *l*<sub class="lower-index">*i*</sub> &lt; *r*<sub class="lower-index">*i*</sub> ≤ *n*). The answer to the query *l*<sub class="lower-index">*i*</sub>, *r*<sub class="lower-index">*i*</sub> is the number of such integers *i* (*l*<sub class="lower-index">*i*</sub> ≤ *i* &lt; *r*<sub class="lower-index">*i*</sub>), that *s*<sub class="lower-index">*i*</sub> = *s*<sub class="lower-index">*i* + 1</sub>.

Ilya the Lion wants to help his friends but is there anyone to help him? Help Ilya, solve the problem.
The first line contains string *s* of length *n* (2 ≤ *n* ≤ 10<sup class="upper-index">5</sup>). It is guaranteed that the given string only consists of characters "." and "#".

The next line contains integer *m* (1 ≤ *m* ≤ 10<sup class="upper-index">5</sup>) — the number of queries. Each of the next *m* lines contains the description of the corresponding query. The *i*-th line contains integers *l*<sub class="lower-index">*i*</sub>, *r*<sub class="lower-index">*i*</sub> (1 ≤ *l*<sub class="lower-index">*i*</sub> &lt; *r*<sub class="lower-index">*i*</sub> ≤ *n*).
s = input()
if s.count('.')==len(s) or s.count('#')==len(s):
    for i in range(int(input())):
        n, m = [int(j)-1 for j in input().split()]
        print(abs(n-m))
else:
    a = s.replace('.#', '')
    if a=='' or a=='.' or a=='#':
        for i in range(int(input())):
            n, m = [int(j)-1 for j in input().split()]
            print(0)
    else:
        for i in range(int(input())):
            n, m = [int(j)-1 for j in input().split()]
            a = s[n:m+1]
            count = 0
            for l in range(len(a)-1):
                if a[l]==a[l+1]:
                    count+=1
        print(count)