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 open, close, sum;
  node() {}
  node(int a, int b, int c) : open(a), close(b), sum(c) {}
};
string s;
node tree[4 * 1123456];
node combine(node a, node b) {
  node ans;
  int inc = min(a.open, b.close);
  ans.sum = a.sum + b.sum + inc;
  ans.open = a.open + b.open - inc;
  ans.close = a.close + b.close - inc;
  return ans;
}
void build(int root, int l, int r) {
  if (l == r) {
    if (s[l] == '(')
      tree[root] = node(1, 0, 0);
    else
      tree[root] = node(0, 1, 0);
    return;
  }
  int m = (l + r) / 2;
  build(2 * root, l, m);
  build(2 * root + 1, m + 1, r);
  tree[root] = combine(tree[2 * root], tree[2 * root + 1]);
}
node query(int root, int l, int r, int a, int b) {
  if (l == a && r == b) return tree[root];
  int m = (l + r) / 2;
  if (b <= m)
    return query(2 * root, l, m, a, b);
  else if (m < a)
    return query(2 * root + 1, m + 1, r, a, b);
  node left = query(2 * root, l, m, a, m);
  node right = query(2 * root + 1, m + 1, r, m + 1, b);
  node ans = combine(left, right);
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> s;
  build(1, 0, s.size() - 1);
  int q;
  cin >> q;
  while (q--) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    cout << 2 * query(1, 0, s.size() - 1, a, b).sum << endl;
  }
}
