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 «()(())(())».
import sys

input = sys.stdin.readline

s = input()
n = len(s)

# Max size of tree
tree = [0] * (2 * n)
closed = [0] * (2 * n)
opened = [0] * (2 * n)


# function to build the tree
def build(arr):
    # insert leaf nodes in tree
    for i in range(n):
        tree[n + i] = 0
        closed[i + n] = int(s[i] == ')')
        opened[i + n] = int(s[i] == '(')
    # print(tree)

    # build the tree by calculating parents
    for i in range(n - 1, 0, -1):
        to_add = min(closed[i << 1 | 1], opened[i << 1])
        tree[i] = tree[i << 1] + tree[i << 1 | 1] + 2 * to_add
        closed[i] = closed[i << 1] + closed[i << 1 | 1] - to_add
        opened[i] = opened[i << 1] + opened[i << 1 | 1] - to_add
        # print(i, closed[i], opened[i], tree[i])


def query(l, r):
    left = []
    right = []
    l += n
    r += n
    while l <= r:
        if (l & 1):
            left.append((tree[l], opened[l], closed[l]))
            l += 1
        if not (r & 1):
            right.append((tree[r], opened[r], closed[r]))
            r -= 1
        l >>= 1
        r >>= 1
    a1 = b1 = c1 = 0
    for a2, b2, c2 in left + right[::-1]:
        to_match = min(b1, c2)
        a1 += a2 + 2 * to_match
        b1 += b2 - to_match
        c1 += c2 - to_match
    return a1


build(s)
# print(query(1, 5))
for _ in range(int(input())):
    a, b = map(int, input().split())
    print(query(a - 1, b - 1));
