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:
if __name__ =="__main__":
    '''s = input()
    n = int(input())

    lr = []
    for i in range(0, n):
        lr.append(input().split())
        lr[i][0] = int(lr[i][0])
        lr[i][1] = int(lr[i][1])

    for i in range(0, n):
        count = 0
        for j in range(lr[i][0]-1, lr[i][1]-1):
            if s[j] == s[j+1]:
                count += 1
        print(count)'''
    #time limiting

    s = input()
    S = [0 for i in range(0,len(s))]
    n = int(input())

    if s[0] == s[1]:
        S[0] = 1
    else:
        S[1] = 0
    for i in range(1, len(s)):
        if s[i] == s[i-1]:
            S[i] = S[i-1] + 1
        else:
            S[i] = S[i-1]

    #make dp S

    lr = []
    for i in range(0, n):
        lr.append(input().split())
        lr[i][0] = int(lr[i][0])
        lr[i][1] = int(lr[i][1])

    for i in range(0, n):
        print(S[lr[i][1]-1] - S[lr[i][0]-1])