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 SegTree {
 private:
  int n;
  vector<int> a, b, c;
  string arr;

 public:
  SegTree(int num, string ar) {
    arr = ar;
    n = num;
    a.resize(4 * n, 0);
    b.resize(4 * n, 0);
    c.resize(4 * n, 0);
    build(1, 0, n - 1);
  }
  void build(int node, int l, int r) {
    if (l == r) {
      a[node] = 0;
      b[node] = (arr[(l = r)] == '(') ? 1 : 0;
      c[node] = (arr[(l = r)] == ')') ? 1 : 0;
      return;
    }
    build(2 * node, l, (l + r) / 2);
    build(2 * node + 1, (l + r) / 2 + 1, r);
    int t = min(b[2 * node], c[2 * node + 1]);
    a[node] = a[2 * node] + a[2 * node + 1] + 2 * t;
    b[node] = b[2 * node] + b[2 * node + 1] - t;
    c[node] = c[2 * node] + c[2 * node + 1] - t;
    return;
  }
  pair<int, pair<int, int> > query(int node, int l, int r, int al, int ar) {
    if (al <= l && r <= ar)
      return make_pair(a[node], make_pair(b[node], c[node]));
    if (r < al || ar < l) return make_pair(0, make_pair(0, 0));
    pair<int, pair<int, int> > left = query(2 * node, l, (l + r) / 2, al, ar);
    pair<int, pair<int, int> > right =
        query(2 * node + 1, (l + r) / 2 + 1, r, al, ar);
    int t = min(left.second.first, right.second.second);
    return make_pair(left.first + right.first + 2 * t,
                     make_pair(left.second.first + right.second.first - t,
                               left.second.second + right.second.second - t));
  }
};
int main() {
  string s;
  cin >> s;
  int q, a, b;
  SegTree st = SegTree(s.length(), s);
  scanf("%d", &q);
  for (int i = 0; i <= q - 1; ++i) {
    scanf("%d", &a);
    scanf("%d", &b);
    printf("%d\n", st.query(1, 0, s.length() - 1, a - 1, b - 1));
  }
  return 0;
}
