Sereja has a bracket sequence s1, s2, ..., sn, or, in other words, a string s of length n, consisting of characters "(" and ")".

Sereja needs to answer m queries, each of them is described by two integers li, ri (1 ≤ li ≤ ri ≤ n). The answer to the i-th query is the length of the maximum correct bracket subsequence of sequence sli, sli + 1, ..., sri. Help Sereja answer all queries.

You can find the definitions for a subsequence and a correct bracket sequence in the notes.

Input

The first line contains a sequence of characters s1, s2, ..., sn (1 ≤ n ≤ 106) without any spaces. Each character is either a "(" or a ")". The second line contains integer m (1 ≤ m ≤ 105) — the number of queries. Each of the next m lines contains a pair of integers. The i-th line contains integers li, ri (1 ≤ li ≤ ri ≤ n) — the description of the i-th query.

Output

Print the answer to each question on a single line. Print the answers in the order they go in the input.

Examples

Input

())(())(())(
7
1 1
2 3
1 2
1 12
8 12
5 11
2 10


Output

0
0
2
10
4
6
6

Note

A subsequence of length |x| of string s = s1s2... s|s| (where |s| is the length of string s) is string x = sk1sk2... sk|x| (1 ≤ k1 < k2 < ... < k|x| ≤ |s|).

A correct bracket sequence is a bracket sequence that can be transformed into a correct aryphmetic expression by inserting characters "1" and "+" between the characters of the string. For example, bracket sequences "()()", "(())" are correct (the resulting expressions "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.

For the third query required sequence will be «()».

For the fourth query required sequence will be «()(())(())».
def main():
    from sys import stdin,stdout
    from collections import deque
    rr,w,it,mp,st = stdin.readline,stdout.write,int,map,str

    arr = stdin.next().rstrip()
    n = len(arr)
    q = it(stdin.next())
    queries = [[] for _ in xrange(n+1)]
    ans = [0]*q

    #Build
    bit = [0]*(n+1)
    for idx,line in enumerate(stdin):
        l,r = mp(it,line.split())
        queries[r-1].append((l-1,r-1,idx))

    stack = deque()
    for j in xrange(n):
        if arr[j] == '(':
            stack.appendleft(j)
        else:
            if stack:
                i = stack.popleft() + 1
                while i <= n: bit[i] += 1; i += (i & (-i))

        for l,r,idx in queries[j]:
            i,k = j+1,l
            s1 = s2 = 0
            while i>0 or k>0:
                if i > 0: s1 += bit[i]; i -= i & (-i)
                if k > 0: s2 += bit[k]; k -= k & (-k)
            ans[idx] = (s1-s2) << 1
            
    w('\n'.join(mp(st,ans)))

main()
