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 = 1e6 + 10;
int n, q;
string s;
struct node {
  int opn, cls, ans;
} seg[4 * maxn];
node ins(int x1, int x2, int x3) {
  node x;
  x.opn = x1, x.cls = x2, x.ans = x3;
  return x;
}
node update(int id, node x, node y) {
  int ch1 = id * 2 + 0, ch2 = id * 2 + 1;
  node p = ins(0, 0, 0);
  p.ans = x.ans + y.ans + min(x.opn, y.cls);
  p.opn = y.opn + (min(x.opn, y.cls) == x.opn ? 0 : x.opn - y.cls);
  p.cls = x.cls + (min(x.opn, y.cls) == y.cls ? 0 : y.cls - x.opn);
  return p;
}
void modify(int l = 0, int r = n, int id = 1) {
  if (r - l == 1) {
    if (s[l] == '(')
      seg[id].opn = 1;
    else
      seg[id].cls = 1;
    return;
  }
  int mid = (l + r) / 2;
  modify(l, mid, id * 2 + 0);
  modify(mid, r, id * 2 + 1);
  seg[id] = update(id, seg[id * 2 + 0], seg[id * 2 + 1]);
  return;
}
node get(int first, int second, int l = 0, int r = n, int id = 1) {
  if (r <= first || second <= l) return ins(0, 0, 0);
  if (first <= l && r <= second)
    return ins(seg[id].opn, seg[id].cls, seg[id].ans);
  int mid = (l + r) / 2;
  node x = get(first, second, l, mid, id * 2 + 0);
  node y = get(first, second, mid, r, id * 2 + 1);
  return update(id, x, y);
}
int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(false);
  cin >> s >> q;
  n = s.size();
  modify();
  for (int i = 0; i < q; i++) {
    int first, second;
    cin >> first >> second;
    first--;
    cout << 2 * get(first, second).ans << '\n';
  }
  return 0;
}
