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 «()(())(())».
#!/usr/bin/env python
# -*- coding: utf-8 -*-
#
#  sem título.py
#  
#  Copyright 2020 Alencar <Alencar@ALENCAR-PC>
#  
#  This program is free software; you can redistribute it and/or modify
#  it under the terms of the GNU General Public License as published by
#  the Free Software Foundation; either version 2 of the License, or
#  (at your option) any later version.
#  
#  This program is distributed in the hope that it will be useful,
#  but WITHOUT ANY WARRANTY; without even the implied warranty of
#  MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
#  GNU General Public License for more details.
#  
#  You should have received a copy of the GNU General Public License
#  along with this program; if not, write to the Free Software
#  Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston,
#  MA 02110-1301, USA.
#  
#  

import os
from sys import stdin
 
import sys
 
 
def solution():
    sys.stdout = os.fdopen(sys.stdout.fileno(), 'wb')
    a = [0]
    for x in stdin.next().rstrip('\r\n'):
        a.append(a[-1] + (1 if x == '(' else -1))
    stdin.next()
    tree = [a]
    while len(a) >= 2:
        a = [min(a[i], a[i + 1]) for i in xrange(0, len(a) - 1, 2)]
        tree.append(a)
    a = tree[0]
    for line in stdin:
        left, right = map(int, line.split())
        left -= 1
 
        ans = 1000 * 1000 * 1000
        l = left
        r = right + 1
        for t in tree:
            if l & 1:
                ans = min(ans, t[l])
                l += 1
            if r & 1:
                r -= 1
                ans = min(ans, t[r])
            l /= 2
            r /= 2
            if l == r:
                break
        ans = right - left - (a[left] - ans) - (a[right] - ans)
        sys.stdout.write('%d\n' % ans)
 
 
solution()
	  				  	    							  			 	  	