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 oo = 1e9 + 1;
const int mod = 1e9 + 7;
const int maxn = 1e6 + 5;
int n, m, u, v;
string s;
struct Node {
  int optimal, close, open;
};
Node St[4 * maxn], res;
Node Join(const Node &x, const Node &y) {
  Node kq;
  kq.optimal = x.optimal + y.optimal + min(x.open, y.close) * 2;
  kq.close = x.close;
  kq.open = y.open;
  if (x.open > y.close)
    kq.open += x.open - y.close;
  else
    kq.close += y.close - x.open;
  return kq;
}
void Init() {
  cin >> s;
  n = s.length();
  s.insert(0, " ");
}
void Build(int l, int r, int id) {
  if (l == r) {
    St[id].optimal = 0;
    if (s[l] == ')')
      St[id].close = 1;
    else
      St[id].open = 1;
    return;
  }
  int mid = (l + r) / 2;
  Build(l, mid, id * 2);
  Build(mid + 1, r, id * 2 + 1);
  St[id] = Join(St[id * 2], St[id * 2 + 1]);
}
void Query(int l, int r, int id, int u, int v) {
  if (u > r || l > v) return;
  if (u <= l && r <= v) {
    res = Join(res, St[id]);
    return;
  }
  int mid = (l + r) / 2;
  Query(l, mid, id * 2, u, v);
  Query(mid + 1, r, id * 2 + 1, u, v);
}
void Solve() {
  cin >> m;
  for (int i = 1; i <= m; ++i) {
    cin >> u >> v;
    res.optimal = res.open = res.close = 0;
    Query(1, n, 1, u, v);
    cout << res.optimal << endl;
  }
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  Init();
  Build(1, n, 1);
  Solve();
}
