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*).
def get_input():
    line = input()
    queries = []
    for _ in range(int(input())):
        queries.append([int(number) - 1 for number in input().split(' ')])
    return line, queries


if __name__ == '__main__':
    line, queries = get_input()
    diffs = [0 if line[i] != line[i + 1] else 1 for i in range(len(line) - 1)]
    print(line)
    print(queries)
    print(diffs)
    pref_sums = [0]
    for i in range(len(diffs)):
        pref_sums.append(pref_sums[i] + diffs[i])
    print(pref_sums)
    for query in queries:
        print(pref_sums[query[1]] - pref_sums[query[0]])

