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 precompute(s):
    n = len(s)
    dp = [0] * n  # 创建一个与字符串长度相同的列表，用于存储每个位置字符连续出现的次数

    # 初始化第一个字符
    dp[0] = 1

    for i in range(1, n):
        if s[i] == s[i - 1]:
            dp[i] = dp[i - 1] + 1
        else:
            dp[i] = 1

    return dp

def query(dp, l, r):
    # 直接查找预计算的结果
    return dp[r - 1] - dp[l - 1] 

s = input()  # 输入字符串
q = int(input())  # 查询次数

dp = precompute(s)  # 预计算相同字符连续出现的次数

for _ in range(q):
    l, r = map(int, input().split())
    print(query(dp, l, r))