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;
const int maxn = 1e6 + 85 - 69;
pair<int, pair<int, int> > seg[maxn * 4];
string s;
int m;
pair<int, pair<int, int> > make(int first = 0, int second = s.size(),
                                int id = 1) {
  if (second - first == 1) {
    if (s[first] == '(')
      seg[id].second.first = 1;
    else
      seg[id].second.second = 1;
    return seg[id];
  }
  int mid = (first + second) >> 1;
  pair<int, pair<int, int> > inter1, inter2;
  inter1 = make(first, mid, id * 2 + 0);
  inter2 = make(mid, second, id * 2 + 1);
  int match = min(inter1.second.first, inter2.second.second);
  seg[id].first = inter1.first + inter2.first + match;
  seg[id].second.first = inter1.second.first + inter2.second.first - match;
  seg[id].second.second = inter1.second.second + inter2.second.second - match;
  return seg[id];
}
pair<int, pair<int, int> > get(int l, int r, int first = 0,
                               int second = s.size(), int id = 1) {
  if (l == first and r == second) return seg[id];
  int mid = (first + second) >> 1;
  pair<int, pair<int, int> > inter1 = make_pair(0, make_pair(0, 0)),
                             inter2 = make_pair(0, make_pair(0, 0)), here;
  if (l < mid) inter1 = get(l, min(mid, r), first, mid, id * 2 + 0);
  if (r > mid) inter2 = get(max(l, mid), r, mid, second, id * 2 + 1);
  int match = min(inter1.second.first, inter2.second.second);
  here.first = inter1.first + inter2.first + match;
  here.second.first = inter1.second.first + inter2.second.first - match;
  here.second.second = inter1.second.second + inter2.second.second - match;
  return here;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> s;
  make();
  cin >> m;
  while (m--) {
    int fi, se;
    cin >> fi >> se;
    cout << 2 * get(fi - 1, se).first << '\n';
  }
  return 0;
}
