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 long long int mod = 1000000007;
const long long int inf = 0x3f3f3f3f3f3f3f3fLL;
const double eps = 1e-9;
const double PI = acos(-1);
long long int i, j, k;
void solve(void);
inline bool equal(double a, double b) { return fabs(a - b) < eps; }
int main() {
  cout << fixed << setprecision(10);
  int t = 1;
  while (t--) solve();
  return 0;
}
string s;
pair<int, pair<long long int, long long int> > segt[4 * 1000005];
void build(int v, int l, int r) {
  if (l == r) {
    segt[v].first = 0;
    segt[v].second.first = (s[l] == '(' ? 1 : 0);
    segt[v].second.second = (s[l] == ')' ? 1 : 0);
    return;
  }
  int m = (l + r) >> 1;
  build(2 * v, l, m);
  build(2 * v + 1, m + 1, r);
  int min = ((segt[2 * v].second.first) < (segt[2 * v + 1].second.second)
                 ? (segt[2 * v].second.first)
                 : (segt[2 * v + 1].second.second));
  segt[v].first = segt[2 * v].first + segt[2 * v + 1].first + 2 * min;
  segt[v].second.first =
      segt[2 * v].second.first + segt[2 * v + 1].second.first - min;
  segt[v].second.second =
      segt[2 * v].second.second + segt[2 * v + 1].second.second - min;
}
pair<int, pair<long long int, long long int> > query(int v, int l, int r,
                                                     int qs, int qe) {
  if (r < qs || l > qe) return {0, {0, 0}};
  if (qs <= l && r <= qe) return segt[v];
  int m = (l + r) >> 1;
  pair<int, pair<long long int, long long int> > p1 =
      query(2 * v, l, m, qs, qe);
  pair<int, pair<long long int, long long int> > p2 =
      query(2 * v + 1, m + 1, r, qs, qe);
  pair<int, pair<long long int, long long int> > ans;
  int min = ((p1.second.first) < (p2.second.second) ? (p1.second.first)
                                                    : (p2.second.second));
  ans.first = p1.first + p2.first + 2 * min;
  ans.second.first = p1.second.first + p2.second.first - min;
  ans.second.second = p1.second.second + p2.second.second - min;
  return ans;
}
void solve() {
  string t;
  cin >> t;
  s = '#';
  s += t;
  int q;
  cin >> q;
  build(1, 1, (int)t.size());
  while (q--) {
    long long int l, r;
    cin >> l >> r;
    cout << query(1, 1, (int)t.size(), l, r).first << '\n';
  }
}
