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 PYTHON3):

Solution:
from sys import stdin, stdout
cin = stdin.readline
cout = stdout.write
mp = lambda:list(map(int, cin().split()))

s = cin()[:-1]
m, = mp()

l = [0]

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

#temp = l.count(1)		
for i in range(m):
	p, r = mp()
	#if r-p >= len(s)//2:
		#cout(str(temp - l[:p].count(1) - l[r:].count(1)) + '\n')
		
	#else: cout(str(l[p:r].count(1)) + '\n')
	cout(str(l[r-1]-l[p-1]) + '\n')