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;
char s[1000010];
int o[4 * 1000010], c[4 * 1000010], t[4 * 1000010];
int n;
void build(int id = 1, int l = 0, int r = n) {
  if (r - l < 2) {
    if (s[l] == '(')
      o[id] = 1;
    else
      c[id] = 1;
    return;
  }
  int mid = (l + r) / 2;
  build(2 * id, l, mid);
  build(2 * id + 1, mid, r);
  int tmp = min(o[2 * id], c[2 * id + 1]);
  t[id] = t[2 * id] + t[2 * id + 1] + tmp;
  o[id] = o[2 * id] + o[2 * id + 1] - tmp;
  c[id] = c[2 * id] + c[2 * id + 1] - tmp;
}
pair<int, pair<int, int> > query(int x, int y, int id = 1, int l = 0,
                                 int r = n) {
  if (l >= y || x >= r)
    return pair<int, pair<int, int> >(0, pair<int, int>(0, 0));
  if (x <= l && r <= y)
    return pair<int, pair<int, int> >(t[id], pair<int, int>(o[id], c[id]));
  int mid = (l + r) / 2;
  pair<int, pair<int, int> > a = query(x, y, 2 * id, l, mid),
                             b = query(x, y, 2 * id + 1, mid, r);
  int T, temp, o, c;
  temp = min(a.second.first, b.second.second);
  T = a.first + b.first + temp;
  o = a.second.first + b.second.first - temp;
  c = a.second.second + b.second.second - temp;
  return pair<int, pair<int, int> >(T, pair<int, int>(o, c));
}
int main() {
  string p;
  cin >> p;
  n = p.length();
  for (int i = 0; i < n; i++) s[i] = p[i];
  build();
  int num_queries;
  scanf("%d", &num_queries);
  for (int i = 0; i < num_queries; i++) {
    int l, r;
    scanf("%d %d", &l, &r);
    int ans = 2 * query(l - 1, r).first;
    cout << ans << "\n";
  }
  return 0;
}
