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*).
import sys
input = sys.stdin.readline

############ ---- Input Functions ---- ############
def inp():
    return(int(input()))
def inlt():
    return(list(map(int,input().split())))
def insr():
    s = input()
    return(list(s[:len(s) - 1]))
def invr():
    return(map(int,input().split()))
############ ---- Input Functions ---- ############

def Ilya_and_Queries():
    string1 = insr()
    n = len(string1)
    binary_converted = []

    for char in string1:
        if char == '.':
            binary_converted.append(0)
        else:
            binary_converted.append(1)
    
    diff = []
    index = 0
    while index <= (n-2):
        diff.append(binary_converted[index] - binary_converted[index+1])
        index += 1
    
    m = inp()
    outputStr = ''
    for i in range(m):
        li,ri = invr()
        sliced_diff = diff[(li-1):(ri-1)]
        res = str(sliced_diff.count(0))
        outputStr += res + '\n'

    outputStr = outputStr.strip() 
    print(outputStr)    
    return

Ilya_and_Queries()