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 + 5;
const int mod = 1e9 + 7;
string str;
struct node {
  int open, close, cur;
  node() { open = close = cur = 0; }
} segtree[4 * N];
inline node join(const node& l, const node& r) {
  node ans;
  ans.cur = l.cur + r.cur;
  ans.open = l.open + r.open;
  ans.close = l.close + r.close;
  int matching = min(l.open, r.close);
  ans.open -= matching;
  ans.close -= matching;
  ans.cur = ans.cur + 2 * matching;
  return ans;
}
void build(int idx, int l, int r) {
  if (l == r) {
    if (str[l] == '(')
      segtree[idx].open++;
    else
      segtree[idx].close++;
    return;
  }
  int mid = (l + r) >> 1;
  build(2 * idx, l, mid);
  build(2 * idx + 1, mid + 1, r);
  segtree[idx] = join(segtree[2 * idx], segtree[2 * idx + 1]);
}
node query(int idx, int l, int r, int ql, int qr) {
  if (ql <= l && r <= qr) return segtree[idx];
  if (ql > r || qr < l) return node();
  int mid = (l + r) >> 1;
  return join(query(2 * idx, l, mid, ql, qr),
              query(2 * idx + 1, mid + 1, r, ql, qr));
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin.exceptions(cin.failbit);
  cin >> str;
  int n = str.size();
  build(1, 0, n - 1);
  int q, l, r;
  cin >> q;
  while (q-- > 0) {
    cin >> l >> r;
    cout << query(1, 0, n - 1, l - 1, r - 1).cur << "\n";
  }
}
