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;
pair<int, int> tree[3 * 1000009];
string str;
char lp = ')';
void init(int node, int st, int end) {
  if (st == end) {
    if (str[st] == lp) {
      tree[node].first = 1;
      tree[node].second = 0;
    } else {
      tree[node].first = 0;
      tree[node].second = 1;
    }
    return;
  }
  int mid = (st + end) / 2;
  init(node * 2, st, mid);
  init(node * 2 + 1, mid + 1, end);
  int l1 = tree[node * 2].first;
  int r1 = tree[node * 2].second;
  int l2 = tree[node * 2 + 1].first;
  int r2 = tree[node * 2 + 1].second;
  tree[node].first = l1 + max(l2 - r1, 0);
  tree[node].second = r2 + max(r1 - l2, 0);
}
pair<int, int> query(int node, int st, int end, int qs, int qe) {
  if (qs > end || qe < st) return make_pair(0, 0);
  if (qs <= st && qe >= end) return tree[node];
  int mid = (st + end) / 2;
  pair<int, int> a = query(2 * node, st, mid, qs, qe);
  pair<int, int> b = query(2 * node + 1, mid + 1, end, qs, qe);
  pair<int, int> c;
  c.first = max(b.first - a.second, 0) + a.first;
  c.second = max(a.second - b.first, 0) + b.second;
  return c;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> str;
  init(1, 0, str.length() - 1);
  int M;
  cin >> M;
  int li, ri;
  for (int i = 0; i < M; i++) {
    cin >> li >> ri;
    pair<int, int> t = query(1, 0, str.length() - 1, li - 1, ri - 1);
    cout << ri - li + 1 - t.first - t.second << endl;
  }
  return 0;
}
