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;
long long n, l, r, q;
string a;
vector<pair<long long, pair<long long int, long long int>>> seg(4 * 1000005);
void build(long long cur, long long st, long long end) {
  if (st == end) {
    seg[cur].first = 0;
    if (a[st] == '(') {
      seg[cur].second.first = 1;
      seg[cur].second.second = 0;
    } else {
      seg[cur].second.first = 0;
      seg[cur].second.second = 1;
    }
    return;
  }
  long long mid = (st + end) >> 1;
  build(2 * cur, st, mid);
  build(2 * cur + 1, mid + 1, end);
  long long mini =
      min(seg[2 * cur].second.first, seg[2 * cur + 1].second.second);
  seg[cur].first = seg[2 * cur].first + seg[2 * cur + 1].first + 2 * mini;
  seg[cur].second.second =
      seg[2 * cur].second.second + seg[2 * cur + 1].second.second - mini;
  seg[cur].second.first =
      seg[2 * cur].second.first + seg[2 * cur + 1].second.first - mini;
}
pair<long long, pair<long long int, long long int>> query(
    long long cur, long long st, long long end, long long l, long long r) {
  if (l <= st && r >= end) return seg[cur];
  if (r < st || l > end) return {0, {0, 0}};
  long long mid = (st + end) >> 1;
  pair<long long, pair<long long int, long long int>> ans1 =
      query(2 * cur, st, mid, l, r);
  pair<long long, pair<long long int, long long int>> ans2 =
      query(2 * cur + 1, mid + 1, end, l, r);
  long long mini = min(ans1.second.first, ans2.second.second);
  pair<long long, pair<long long int, long long int>> ans;
  ans.first = ans1.first + ans2.first + 2 * mini;
  ans.second.second = ans1.second.second + ans2.second.second - mini;
  ans.second.first = ans1.second.first + ans2.second.first - mini;
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) {
    string s;
    cin >> s;
    a = '*' + s;
    n = s.size();
    build(1, 1, n);
    cin >> q;
    for (long long int i = 0; i < q; i++) {
      cin >> l >> r;
      cout << query(1, 1, n, l, r).first << '\n';
    }
  }
  return 0;
}
