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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  using tiii = tuple<int, int, int>;
  struct ST {
    vector<int> balanced, open, close;
    ST(int n) {
      balanced.resize(4 * n + 5);
      open.resize(4 * n + 5);
      close.resize(4 * n + 5);
      return;
    }
    void upd(int i, int l, int r, int pos, char c) {
      if (pos < l || r < pos) return;
      if (l == r) {
        if (c == '(')
          open[i] = 1;
        else
          close[i] = 1;
        if (pos == 4) {
        }
        return;
      }
      int mid = (l + r) >> 1;
      if (pos <= mid)
        upd(i * 2, l, mid, pos, c);
      else
        upd(i * 2 + 1, mid + 1, r, pos, c);
      balanced[i] = balanced[i * 2] + balanced[i * 2 + 1] +
                    min(open[i * 2], close[i * 2 + 1]) * 2;
      open[i] =
          open[i * 2] + open[i * 2 + 1] - min(open[i * 2], close[i * 2 + 1]);
      close[i] =
          close[i * 2] + close[i * 2 + 1] - min(open[i * 2], close[i * 2 + 1]);
      return;
    }
    tiii getbalanced(int i, int l, int r, int u, int v) {
      if (v < l || r < u || u > v) {
        return make_tuple(0, 0, 0);
      }
      if (u <= l && r <= v) return make_tuple(balanced[i], open[i], close[i]);
      int mid = (l + r) >> 1;
      tiii t1 = getbalanced(i * 2, l, mid, u, v);
      tiii t2 = getbalanced(i * 2 + 1, mid + 1, r, u, v);
      return make_tuple(
          get<0>(t1) + get<0>(t2) + min(get<1>(t1), get<2>(t2)) * 2,
          get<1>(t1) + get<1>(t2) - min(get<1>(t1), get<2>(t2)),
          get<2>(t1) + get<2>(t2) - min(get<1>(t1), get<2>(t2)));
    }
  };
  string s;
  cin >> s;
  int n = s.size();
  ST tree(n);
  s = ' ' + s;
  for (int i = 1; i <= n; i++) tree.upd(1, 1, n, i, s[i]);
  int q;
  cin >> q;
  while (q--) {
    int l, r;
    cin >> l >> r;
    cout << get<0>(tree.getbalanced(1, 1, n, l, r)) << '\n';
  }
  return 0;
}
