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:
import math
import bisect
import sys
from collections import OrderedDict
input = sys.stdin.readline
def inp():
	return(int(input()))
def inlt():
	return(list(map(int,input().split())))
def insr():
	s = input()
	return(s[:len(s)-1])
def invr():
	return(map(int,input().split()))
# T = inp()
# for t in range(T):
# N, M = invr()
# A = inlt()
S = insr()
cum = []
prev = 0
for i in range(len(S)-1):
	if S[i] == S[i+1]:
		cum.append(prev+1)
		prev += 1
	else:
		cum.append(prev)
cum.append(prev)
M = inp()
# print(cum)
for m in range(M):
	l, r = invr()
	l-= 1
	r -= 1
	# if l >= 1:
	# print(l)
	# print(r)
	if l >0 :
		print(cum[r-1]-cum[l-1])
	else:
		print(cum[r-1]-0)
	# print(cum[r-1])
	# print(cum[l-1])
	# else:
	# 	print(cum[r-1]-cum[l])