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;
unsigned long long gcd(unsigned long long a, unsigned long long b) {
  return (b == 0 ? a : gcd(b, a % b));
}
long long n, m, x, y, r, z, l, k;
string s;
long long max_size;
long long segarr[1000001];
long long* bal;
long long* op;
long long* clo;
pair<long long, pair<long long, long long> > query(long long ss, long long se,
                                                   long long si, long long deb,
                                                   long long fin) {
  if (se < deb || ss > fin) {
    return {0, {0, 0}};
  }
  if (deb <= ss && fin >= se) {
    return {bal[si], {op[si], clo[si]}};
  }
  long long mid = ss + (se - ss) / 2;
  pair<long long, pair<long long, long long> > left, right;
  left = query(ss, mid, si * 2 + 1, deb, fin);
  right = query(mid + 1, se, si * 2 + 2, deb, fin);
  long long tmp = min(left.second.first, right.second.second);
  return {left.first + right.first + tmp,
          {left.second.first + right.second.first - tmp,
           left.second.second + right.second.second - tmp}};
}
void build(long long ss, long long se, long long si) {
  if (ss == se) {
    clo[si] = 0;
    op[si] = 0;
    bal[si] = 0;
    if (s[ss] == ')')
      clo[si] = 1;
    else
      op[si] = 1;
    return;
  }
  long long mid = ss + (se - ss) / 2;
  build(ss, mid, si * 2 + 1);
  build(mid + 1, se, si * 2 + 2);
  long long tmp = min(op[si * 2 + 1], clo[si * 2 + 2]);
  bal[si] = tmp + bal[si * 2 + 1] + bal[si * 2 + 2];
  op[si] = op[si * 2 + 1] + op[si * 2 + 2] - tmp;
  clo[si] = clo[si * 2 + 1] + clo[si * 2 + 2] - tmp;
  return;
}
void buildinit() {
  long long x = (long long)(ceil(log2(n)));
  max_size = 2 * (long long)pow(2, x) - 1;
  bal = new long long[max_size];
  op = new long long[max_size];
  clo = new long long[max_size];
  build(0, n - 1, 0);
}
void solve() {
  cin >> s;
  n = s.size();
  buildinit();
  cin >> m;
  while (m--) {
    cin >> x >> y;
    x--;
    y--;
    cout << query(0, n - 1, 0, x, y).first * 2 << "\n";
  }
}
int main() {
  ios::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  solve();
  return 0;
}
