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 tree {
  int opb = 0, clb = 0, len = 0;
  void merge(tree *t1, tree *t2) {
    int t = min(t1->opb, t2->clb);
    this->opb = (t1->opb - t) + t2->opb;
    this->clb = t1->clb + (t2->clb - t);
    this->len = t1->len + t2->len + 2 * t;
  }
};
tree zero;
void update(int l, int r, int b, int e, char c, vector<tree> &Tree) {
  if (e <= l || r <= b) {
  } else if (l <= b && e <= r) {
    (c == '(') ? Tree[b + e].opb = 1 : Tree[b + e].opb = 0;
    Tree[b + e].clb = 1 - Tree[b + e].opb;
    Tree[b + e].len = 0;
  } else {
    update(l, r, b, (b + e) / 2, c, Tree);
    update(l, r, (b + e) / 2, e, c, Tree);
    Tree[b + e].merge(&Tree[b + (b + e) / 2], &Tree[(b + e) / 2 + e]);
  }
}
tree query(int l, int r, int b, int e, vector<tree> &Tree) {
  if (e <= l || r <= b)
    return zero;
  else if (l <= b && e <= r)
    return Tree[b + e];
  else {
    tree t1 = query(l, r, b, (b + e) / 2, Tree),
         t2 = query(l, r, (b + e) / 2, e, Tree), t3;
    t3.merge(&t1, &t2);
    return t3;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  int n, s = 1;
  string c;
  cin >> c;
  n = c.size();
  while (s < 2 * n) s *= 2;
  vector<tree> Tree(s);
  for (int i = 0; i < n; i++) update(i, i + 1, 0, s / 2, c[i], Tree);
  int q, l, r;
  cin >> q;
  for (int h = 0; h < q; h++) {
    cin >> l >> r;
    cout << query(l - 1, r, 0, s / 2, Tree).len << endl;
  }
  return 0;
}
