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;
const int MAXN = 1 << 21;
int END = 1;
struct node {
  int ans, left, right;
};
int n, m;
string s;
node t[MAXN];
int min(int a, int b) { return a < b ? a : b; }
int max(int a, int b) { return a > b ? a : b; }
node merge(node x, node y) {
  node res;
  int comb = min(x.left, y.right);
  res.ans = x.ans + y.ans + comb;
  res.left = x.left + y.left - comb;
  res.right = x.right + y.right - comb;
  return res;
}
void build(int v, int l, int r) {
  if (l == r) {
    if (l <= n) {
      t[v] = {0, (s[l - 1] == '(' ? 1 : 0), (s[l - 1] == ')' ? 1 : 0)};
    } else {
      t[v] = {0, 0, 0};
    }
  } else {
    int h = (l + r) / 2;
    build(v * 2, l, h);
    build(v * 2 + 1, h + 1, r);
    t[v] = merge(t[v * 2], t[v * 2 + 1]);
  }
}
node get(int v, int l, int r, int gl, int gr) {
  if (gl > gr) {
    return {0, 0, 0};
  }
  if (gl == l && gr == r) {
    return t[v];
  } else {
    int h = (l + r) / 2;
    node l_ans = get(v * 2, l, h, gl, min(h, gr));
    node r_ans = get(v * 2 + 1, h + 1, r, max(h + 1, gl), gr);
    return merge(l_ans, r_ans);
  }
}
int main() {
  cin >> s;
  n = s.length();
  while (END < n) {
    END *= 2;
  }
  build(1, 1, END);
  cin >> m;
  for (int i = 0; i < m; i++) {
    int l, r;
    cin >> l >> r;
    cout << get(1, 1, END, l, r).ans * 2 << endl;
  }
  return 0;
}
