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 + 6;
int Q, n;
string str;
struct Node {
  int open, close, valid;
  void merge(Node a, Node b) {
    int t = min(a.open, b.close);
    open = a.open - t + b.open;
    close = a.close + b.close - t;
    valid = a.valid + b.valid + t;
  }
} seg[N << 2];
void build(int s = 0, int e = str.size(), int id = 1) {
  if (e - s == 1) {
    seg[id] = {(str[s] == '('), (str[s] == ')'), 0};
    return;
  }
  int mid = s + e >> 1;
  build(s, mid, id << 1);
  build(mid, e, id << 1 | 1);
  seg[id].merge(seg[id << 1], seg[id << 1 | 1]);
}
Node get(int l, int r, int s = 0, int e = str.size(), int id = 1) {
  if (l <= s && e <= r) return seg[id];
  if (l >= e || r <= s) return {0, 0, 0};
  int mid = e + s >> 1;
  Node res;
  res.merge(get(l, r, s, mid, id << 1), get(l, r, mid, e, id << 1 | 1));
  return res;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  cin >> str >> Q;
  build();
  for (int q = 0, l, r; q < Q; q++) {
    cin >> l >> r;
    Node ans = get(--l, r);
    cout << 2 * ans.valid << endl;
  }
}
