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;
string s;
int n, m;
pair<int, pair<int, int> > st[4000001];
void update(int id, int l, int r) {
  if (l == r) {
    if (s[l] == '(') {
      st[id] = pair<int, pair<int, int> >(0, pair<int, int>(1, 0));
    } else {
      st[id] = pair<int, pair<int, int> >(0, pair<int, int>(0, 1));
    }
    return;
  }
  int mid = (l + r) / 2;
  update(id * 2, l, mid);
  update(id * 2 + 1, mid + 1, r);
  int tmp = min(st[id * 2].second.first, st[id * 2 + 1].second.second);
  st[id].first = st[id * 2].first + st[id * 2 + 1].first + tmp;
  st[id].second.first =
      st[id * 2].second.first + st[id * 2 + 1].second.first - tmp;
  st[id].second.second =
      st[id * 2].second.second + st[id * 2 + 1].second.second - tmp;
}
pair<int, pair<int, int> > get(int id, int l, int r, int u, int v) {
  if (v < l || u > r) {
    return pair<int, pair<int, int> >(0, pair<int, int>(0, 0));
  }
  if (l >= u && r <= v) {
    return st[id];
  }
  int mid = (l + r) / 2;
  pair<int, pair<int, int> > a = get(id * 2, l, mid, u, v);
  pair<int, pair<int, int> > b = get(id * 2 + 1, mid + 1, r, u, v);
  int tmp = min(a.second.first, b.second.second);
  int res = (a.first + b.first + tmp);
  int o = a.second.first + b.second.first - tmp;
  int c = a.second.second + b.second.second - tmp;
  return pair<int, pair<int, int> >(res, pair<int, int>(o, c));
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  n = s.size();
  s = '#' + s;
  update(1, 1, n);
  cin >> m;
  while (m--) {
    int l, r;
    cin >> l >> r;
    cout << get(1, 1, n, l, r).first * 2 << "\n";
  }
}
