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;
struct info {
  int lb, rb, ans;
} seg[4 * 1000001];
string s;
info make_info(int lb_, int rb_, int ans_) {
  info d;
  d.lb = lb_;
  d.rb = rb_;
  d.ans = ans_;
  return d;
}
info combine(info l, info r) {
  int tmp = min(l.lb, r.rb);
  info d;
  d.ans = l.ans + r.ans + tmp;
  d.lb = l.lb + r.lb - tmp;
  d.rb = l.rb + r.rb - tmp;
  return d;
}
void build(int node, int L, int R) {
  if (L == R) {
    seg[node] = s[L - 1] == '(' ? make_info(1, 0, 0) : make_info(0, 1, 0);
    return;
  }
  int mid = (L + R) / 2;
  build(node * 2, L, mid);
  build(node * 2 + 1, mid + 1, R);
  seg[node] = combine(seg[node * 2], seg[node * 2 + 1]);
}
info query(int node, int L, int R, int l, int r) {
  if (l > r) return make_info(0, 0, 0);
  if (L == l && R == r) return seg[node];
  int mid = (L + R) / 2;
  return combine(query(node * 2, L, mid, l, min(r, mid)),
                 query(node * 2 + 1, mid + 1, R, max(l, mid + 1), r));
}
int main() {
  int m, l, r;
  cin >> s >> m;
  int n = s.size();
  build(1, 1, n);
  while (m--) {
    scanf("%d %d", &l, &r);
    printf("%d\n", query(1, 1, n, l, r).ans * 2);
  }
  return 0;
}
