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 = 1234567;
struct Node {
  int open, close, answer;
  Node() : open(0), close(0), answer(0) {}
} seg[4 * N];
string s;
int n;
void merge(Node a, Node b, Node& c) {
  c.open = a.open + b.open;
  c.close = a.close + b.close;
  c.answer = a.answer + b.answer + min(a.open - a.answer, b.close - b.answer);
}
Node merge(Node a, Node b) {
  Node c = Node();
  merge(a, b, c);
  return c;
}
void build(int v = 1, int p = 0, int q = n) {
  if (q - p == 1) {
    seg[v].open = s[p] == '(';
    seg[v].close = s[p] == ')';
    return;
  }
  int mid = (p + q) / 2;
  build(2 * v, p, mid);
  build(2 * v + 1, mid, q);
  merge(seg[2 * v], seg[2 * v + 1], seg[v]);
}
Node ask(int l, int r, int v = 1, int p = 0, int q = n) {
  if (l >= q || r <= p) {
    return Node();
  }
  if (l <= p && r >= q) {
    return seg[v];
  }
  int mid = (p + q) / 2;
  return merge(ask(l, r, 2 * v, p, mid), ask(l, r, 2 * v + 1, mid, q));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> s;
  n = s.length();
  build();
  int q;
  cin >> q;
  while (q--) {
    int l, r;
    cin >> l >> r;
    --l;
    cout << ask(l, r).answer * 2 << '\n';
  }
  return 0;
}
