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 MX = 3e6;
int len[MX], opn[MX], cls[MX];
string s;
pair<int, pair<int, int> > combine(int len1, int len2, int opn1, int opn2,
                                   int cls1, int cls2) {
  int t = min(cls2, opn1);
  return {len1 + len2 + t, {opn1 + opn2 - t, cls1 + cls2 - t}};
}
void build(int pos, int l, int r) {
  if (l == r) {
    len[pos] = 0;
    if (s[l] == '(') {
      opn[pos] = 1;
      cls[pos] = 0;
    }
    if (s[l] == ')') {
      opn[pos] = 0;
      cls[pos] = 1;
    }
    return;
  }
  int mid = (l + r) / 2;
  build(pos * 2, l, mid);
  build(pos * 2 + 1, mid + 1, r);
  len[pos] = combine(len[pos * 2], len[pos * 2 + 1], opn[pos * 2],
                     opn[pos * 2 + 1], cls[pos * 2], cls[pos * 2 + 1])
                 .first;
  opn[pos] = combine(len[pos * 2], len[pos * 2 + 1], opn[pos * 2],
                     opn[pos * 2 + 1], cls[pos * 2], cls[pos * 2 + 1])
                 .second.first;
  cls[pos] = combine(len[pos * 2], len[pos * 2 + 1], opn[pos * 2],
                     opn[pos * 2 + 1], cls[pos * 2], cls[pos * 2 + 1])
                 .second.second;
}
pair<int, pair<int, int> > ans(int pos, int l, int r, int ql, int qr) {
  if (l > qr || r < ql) {
    return {0, {0, 0}};
  }
  if (l >= ql && r <= qr) {
    return {len[pos], {opn[pos], cls[pos]}};
  }
  int mid = (l + r) / 2;
  pair<int, pair<int, int> > a = ans(pos * 2, l, mid, ql, qr),
                             b = ans(pos * 2 + 1, mid + 1, r, ql, qr);
  return combine(a.first, b.first, a.second.first, b.second.first,
                 a.second.second, b.second.second);
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  build(1, 0, s.size() - 1);
  int q;
  cin >> q;
  while (q--) {
    int l, r;
    cin >> l >> r;
    l--;
    r--;
    cout << ans(1, 0, s.size() - 1, l, r).first * 2 << endl;
  }
  return 0;
}
