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 N = 1e6 + 100;
const int M = 1e5 + 100;
string s;
int n;
pair<int, int> a[N];
pair<int, pair<int, int>> seg[4 * N];
pair<int, pair<int, int>> merge(pair<int, pair<int, int>> i,
                                pair<int, pair<int, int>> j) {
  pair<int, pair<int, int>> ret;
  int tmp = min(i.second.second, j.second.first);
  ret.first = i.first + j.first + tmp;
  ret.second.first = i.second.first + j.second.first - tmp;
  ret.second.second = i.second.second - tmp + j.second.second;
  return ret;
}
void build(int id, int l, int r) {
  if (l + 1 == r) {
    if (s[l] == '(')
      seg[id] = {0, {0, 1}};
    else
      seg[id] = {0, {1, 0}};
    return;
  }
  int mid = (r + l) >> 1;
  build(2 * id + 0, l, mid);
  build(2 * id + 1, mid, r);
  seg[id] = merge(seg[2 * id], seg[2 * id + 1]);
}
pair<int, pair<int, int>> get(int id, int l, int r, int st, int en) {
  if (st >= r or en <= l) return {0, {0, 0}};
  if (st <= l and r <= en) return seg[id];
  int mid = (r + l) / 2;
  return merge(get(id * 2, l, mid, st, en), get(id * 2 + 1, mid, r, st, en));
}
int main() {
  cin >> s;
  build(1, 0, s.size());
  cin >> n;
  for (int i = 0; i < n; i++) {
    int u, v;
    cin >> u >> v;
    cout << get(1, 0, s.size(), u - 1, v).first * 2 << '\n';
  }
}
