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 Node {
  int optimal;
  int open;
  int close;
  Node() { optimal = open = close = 0; }
  Node(int a, int b, int c) {
    optimal = a;
    open = b;
    close = c;
  }
};
Node operator+(const Node& left, const Node& right) {
  Node res;
  int tmp = min(left.open, right.close);
  res.optimal = left.optimal + right.optimal + tmp;
  res.open = left.open + right.open - tmp;
  res.close = left.close + right.close - tmp;
  return res;
}
int n, m, l, r;
string s;
Node st[1000001 << 2];
void build(int root, int l, int r) {
  if (l > r) return;
  if (l == r) {
    if (s[l - 1] == '(')
      st[root] = Node(0, 1, 0);
    else
      st[root] = Node(0, 0, 1);
    return;
  }
  int mid = (l + r) >> 1;
  build(root << 1, l, mid);
  build(root << 1 ^ 1, mid + 1, r);
  st[root] = st[root << 1] + st[root << 1 ^ 1];
}
Node get(int root, int l, int r, int u, int v) {
  if (v < l || u > r) return Node(0, 0, 0);
  if (l >= u && r <= v) return st[root];
  int mid = (l + r) >> 1;
  return (get(root << 1, l, mid, u, v) + get(root << 1 ^ 1, mid + 1, r, u, v));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> s;
  n = ((int)(s).size());
  build(1, 1, n);
  cin >> m;
  while (m--) {
    cin >> l >> r;
    cout << get(1, 1, n, l, r).optimal * 2 << endl;
  }
}
