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;
class SegmentTree {
 private:
  vector<int> o, c, s;
  vector<int> open, close;
  int left(int i) { return i << 1; }
  int right(int i) { return (i << 1) + 1; }
  void build(int p, int l, int r) {
    if (l == r) {
      o[p] = open[l];
      c[p] = close[l];
      return;
    }
    build(left(p), l, (l + r) / 2);
    build(right(p), (l + r) / 2 + 1, r);
    int tmp = min(o[left(p)], c[right(p)]);
    s[p] = s[left(p)] + s[right(p)] + tmp;
    o[p] = o[left(p)] + o[right(p)] - tmp;
    c[p] = c[left(p)] + c[right(p)] - tmp;
  }
  pair<int, pair<int, int> > query(int p, int l, int r, int i, int j) {
    if (i > r || j < l)
      return pair<int, pair<int, int> >(0, pair<int, int>(0, 0));
    if (i <= l && j >= r)
      return pair<int, pair<int, int> >(s[p], pair<int, int>(o[p], c[p]));
    pair<int, pair<int, int> > node_left = query(left(p), l, (l + r) / 2, i, j);
    pair<int, pair<int, int> > node_right =
        query(right(p), (l + r) / 2 + 1, r, i, j);
    int tmp = min(node_left.second.first, node_right.second.second);
    return pair<int, pair<int, int> >(
        node_left.first + node_right.first + tmp,
        pair<int, int>(
            node_left.second.first + node_right.second.first - tmp,
            node_left.second.second + node_right.second.second - tmp));
  }

 public:
  SegmentTree(vector<int> &_open, vector<int> &_close) {
    open = _open;
    close = _close;
    o.assign(open.size() * 4, 0);
    c.assign(open.size() * 4, 0);
    s.assign(open.size() * 4, 0);
    build(1, 0, open.size() - 1);
  }
  int query(int i, int j) {
    return query(1, 0, open.size() - 1, i, j).first * 2;
  }
};
int main() {
  string s;
  getline(cin, s);
  vector<int> o(s.size(), 0);
  vector<int> c(s.size(), 0);
  for (int i = 0; i < s.size(); ++i) {
    if (s[i] == '(')
      o[i] = 1;
    else
      c[i] = 1;
  }
  SegmentTree st(o, c);
  int q;
  cin >> q;
  while (q--) {
    int l, r;
    cin >> l >> r;
    l--;
    r--;
    cout << st.query(l, r) << "\n";
  }
  return 0;
}
