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;
int q, uz;
char s[1000005];
pair<int, pair<int, int> > tree[1000005 * 4];
void build(int node, int start, int end) {
  if (start == end) {
    tree[node] = make_pair(0, make_pair(s[start] == '(', s[start] == ')'));
    return;
  }
  build(node * 2, start, (start + end) / 2),
      build(node * 2 + 1, (start + end) / 2 + 1, end);
  int mn = min(tree[node * 2].second.first, tree[node * 2 + 1].second.second);
  tree[node] = make_pair(tree[node * 2].first + tree[node * 2 + 1].first + mn,
                         make_pair(tree[node * 2].second.first +
                                       tree[node * 2 + 1].second.first - mn,
                                   tree[node * 2].second.second +
                                       tree[node * 2 + 1].second.second - mn));
}
pair<int, pair<int, int> > query(int node, int start, int end, int l, int r) {
  if (start > end || start > r || end < l) return make_pair(0, make_pair(0, 0));
  if (start >= l && end <= r) return tree[node];
  pair<int, pair<int, int> > bir =
      query(node * 2, start, (start + end) / 2, l, r);
  pair<int, pair<int, int> > iki =
      query(node * 2 + 1, (start + end) / 2 + 1, end, l, r);
  int mn = min(bir.second.first, iki.second.second);
  return make_pair(bir.first + iki.first + mn,
                   make_pair(bir.second.first + iki.second.first - mn,
                             bir.second.second + iki.second.second - mn));
}
int main() {
  scanf("%s", s + 1);
  uz = strlen(s + 1);
  build(1, 1, uz);
  int q;
  scanf("%d", &q);
  while (q--) {
    int x, y;
    scanf("%d %d", &x, &y);
    printf("%d\n", query(1, 1, uz, x, y).first * 2);
  }
  return 0;
}
