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 $ {
  $() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
  }
} $;
const int Maxn = 1000005;
string s;
struct node {
  int answer, left, right;
  node() : answer(0), left(0), right(0) {}
};
node tree[Maxn << 2];
node merge(node a, node b) {
  node c;
  c.answer = a.answer + b.answer;
  int a1 = a.left, a2 = b.right;
  int m = min(a1, a2);
  c.answer += m;
  a1 -= m, a2 -= m;
  c.left = a1 + b.left;
  c.right = a.right + a2;
  return c;
}
int left(int v) { return (v << 1); }
int right(int v) { return (v << 1 | 1); }
void build(int v, int l, int r) {
  if (l == r) {
    tree[v].answer = 0;
    tree[v].left = (s[l] == '(');
    tree[v].right = (s[l] == ')');
    return;
  }
  int mid = (l + r) / 2;
  build(left(v), l, mid);
  build(right(v), mid + 1, r);
  tree[v] = merge(tree[left(v)], tree[right(v)]);
}
node query(int v, int st, int end, int l, int r) {
  if (r < st || end < l) {
    node temp;
    return temp;
  }
  if (l <= st and end <= r) {
    return tree[v];
  }
  int mid = (st + end) / 2;
  node q1 = query(left(v), st, mid, l, r);
  node q2 = query(right(v), mid + 1, end, l, r);
  return merge(q1, q2);
}
int main() {
  int q, l, r, n;
  cin >> s;
  n = s.length();
  build(1, 0, n - 1);
  cin >> q;
  while (q--) {
    cin >> l >> r;
    l--, r--;
    node ans = query(1, 0, n - 1, l, r);
    cout << (2 * ans.answer) << endl;
  }
  return 0;
}
