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;
using st = string;
template <class S>
using ve = vector<S>;
template <class S>
using gr = greater<S>;
template <class S>
using le = less<S>;
const double pi = 2 * acos(0.0);
const int oo = 0x3f3f3f3f;
const double inf = 1.0 / 0.0;
struct ut {
  int len, a, b;
};
class segtree {
 private:
  ve<ut> v, tree;
  int n;

 public:
  segtree(){};
  void pp(ve<ut>& vf, char c) {
    int l = (c == '(' ? 1 : 0);
    vf.push_back({0, l, 1 - l});
  }
  segtree(st& arr) {
    n = arr.length();
    for (auto& x : arr) pp(v, x);
    tree.assign(4 * n + 4, {0, 0, 0});
    build(1, 0, n - 1);
  }
  bool ni(ut y) { return (!y.len && !y.a && !y.b); }
  ut func(ut x, ut y) {
    if (ni(x)) return y;
    if (ni(y)) return x;
    return {x.len + y.len + min(x.a, y.b), y.a + max(0, x.a - y.b),
            x.b + max(0, y.b - x.a)};
  }
  void build(int node, int l, int r) {
    if (l == r)
      tree[node] = {v[l].len, v[l].a, v[l].b};
    else {
      int m = (l + r) / 2, le = 2 * node, ri = 2 * node + 1;
      build(le, l, m);
      build(ri, m + 1, r);
      tree[node] = func(tree[le], tree[ri]);
    }
  }
  ut query(int node, int l, int r, int ql, int qr) {
    if (ql > qr)
      return {0, 0, 0};
    else if (l == ql && r == qr)
      return tree[node];
    else {
      int m = (l + r) / 2;
      return func(query(2 * node, l, m, ql, min(qr, m)),
                  query(2 * node + 1, m + 1, r, max(m + 1, ql), qr));
    }
  }
  int query(int l, int r) { return 2 * query(1, 0, n - 1, l, r).len; }
  void pr(int index) {
    cout << index << " : {" << tree[index].len << "," << tree[index].a << ","
         << tree[index].b << "}\n";
  }
};
int main() {
  ios_base::sync_with_stdio(0);
  st s;
  cin >> s;
  segtree tr(s);
  int q;
  cin >> q;
  for (__typeof(q) i = (0) - ((0) > (q)); i != (q) - ((0) > (q));
       i += 1 - 2 * ((0) > (q))) {
    int x, y;
    cin >> x >> y;
    cout << tr.query(x - 1, y - 1);
    cout << "\n";
    ;
  }
  return 0;
}
