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 «()(())(())».
#include <bits/stdc++.h>
using namespace std;
const int N = 1e7 + 6;
string s;
long long opene[N], close[N], ans[N], m;
const pair<long long, pair<long long, long long> > zero = {0, {0, 0}};
void segment(long long l, long long r, long long num) {
  if (r - l <= 1) {
    if (s[l] == ')')
      close[num] = 1;
    else
      opene[num] = 1;
    return;
  }
  long long mid = (l + r) / 2;
  segment(l, mid, 2 * num);
  segment(mid, r, 2 * num + 1);
  long long add = min(opene[2 * num], close[2 * num + 1]);
  opene[num] = opene[2 * num] + opene[2 * num + 1] - add;
  close[num] = close[2 * num] + close[2 * num + 1] - add;
  ans[num] = ans[2 * num] + ans[2 * num + 1] + add;
}
pair<long long, pair<long long, long long> > solve(long long x, long long y,
                                                   long long l, long long r,
                                                   long long num) {
  pair<long long, pair<long long, long long> > Ans = zero;
  if (l >= y || x >= r) return zero;
  if (x <= l && r <= y)
    return pair<long long, pair<long long, long long> >(
        ans[num], {opene[num], close[num]});
  long long mid = (l + r) / 2;
  pair<long long, pair<long long, long long> > Left =
      solve(x, y, l, mid, 2 * num);
  pair<long long, pair<long long, long long> > Righ =
      solve(x, y, mid, r, 2 * num + 1);
  long long add = min(Left.second.first, Righ.second.second);
  long long F = add + Left.first + Righ.first;
  long long FF = Righ.second.first + Left.second.first - add;
  long long FS = Righ.second.second + Left.second.second - add;
  Ans = {F, {FF, FS}};
  return Ans;
}
int main(int argc, const char* argv[]) {
  getline(cin, s);
  long long n = (long long)s.length();
  segment(0, n, 1);
  scanf("%I64d", &m);
  for (long long i = 0; i < m; i++) {
    long long c, r;
    scanf("%I64d", &c);
    scanf("%I64d", &r);
    pair<long long, pair<long long, long long> > x = solve(c - 1, r, 0, n, 1);
    printf("%I64d\n", x.first * 2);
  }
  return 0;
}
