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 brace {
  brace() : open(0), closed(0), correct(0) {}
  int open;
  int closed;
  int correct;
};
void build(const string &s, vector<brace> &tree, int v, int tl, int tr) {
  if (tl == tr) {
    tree[v].open = s[tl] == '(';
    tree[v].closed = !tree[v].open;
    tree[v].correct = 0;
  } else {
    int tm = (tl + tr) / 2;
    build(s, tree, v * 2, tl, tm);
    build(s, tree, v * 2 + 1, tm + 1, tr);
    tree[v].open = tree[v * 2].open + tree[v * 2 + 1].open;
    tree[v].closed = tree[v * 2].closed + tree[v * 2 + 1].closed;
    int x = min(tree[v * 2].open - tree[v * 2].correct,
                tree[v * 2 + 1].closed - tree[v * 2 + 1].correct);
    tree[v].correct = tree[v * 2].correct + tree[v * 2 + 1].correct + x;
  }
}
brace request(const vector<brace> &tree, int v, int tl, int tr, int l, int r) {
  if (l > r) return brace();
  if (l == tl && r == tr) return tree[v];
  int tm = (tl + tr) / 2;
  brace bl = request(tree, v * 2, tl, tm, l, min(r, tm));
  brace br = request(tree, v * 2 + 1, tm + 1, tr, max(l, tm + 1), r);
  brace new_br;
  new_br.open = br.open + bl.open;
  new_br.closed = br.closed + bl.closed;
  int x = min(bl.open - bl.correct, br.closed - br.correct);
  new_br.correct = br.correct + bl.correct + x;
  return new_br;
}
int main() {
  string s;
  cin >> s;
  vector<brace> segment_tree(s.size() * 4);
  build(s, segment_tree, 1, 0, s.size() - 1);
  int n;
  cin >> n;
  for (int i = 0; i != n; ++i) {
    int l, r;
    cin >> l >> r;
    --l;
    --r;
    cout << request(segment_tree, 1, 0, s.size() - 1, l, r).correct * 2 << endl;
  }
}
