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;
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-ffloat-store")
#pragma GCC optimize("-fno-defer-pop")
long long int power(long long int a, long long int b, long long int m) {
  if (b == 0) return 1;
  if (b == 1) return a % m;
  long long int t = power(a, b / 2, m) % m;
  t = (t * t) % m;
  if (b & 1) t = ((t % m) * (a % m)) % m;
  return t;
}
long long int modInverse(long long int a, long long int m) {
  return power(a, m - 2, m);
}
string s;
pair<long long int, long long int> st[4 * 1000005];
void build(long long int node, long long int nl, long long int nr) {
  if (nl == nr) {
    if (s[nl - 1] == '(')
      st[node] = {1, 0};
    else
      st[node] = {0, 1};
    return;
  }
  build(2 * node, nl, (nl + nr) / 2);
  build(2 * node + 1, (nl + nr) / 2 + 1, nr);
  st[node] = {st[2 * node + 1].first +
                  max(0ll, st[2 * node].first - st[2 * node + 1].second),
              st[2 * node].second +
                  max(0ll, st[2 * node + 1].second - st[2 * node].first)};
}
pair<long long int, long long int> query(long long int node, long long int nl,
                                         long long int nr, long long int ql,
                                         long long int qr) {
  if (nl > qr || ql > nr) {
    return {0, 0};
  }
  if (nl >= ql && qr >= nr) {
    return st[node];
  }
  pair<long long int, long long int> a, b;
  a = query(2 * node, nl, (nl + nr) / 2, ql, qr);
  b = query(2 * node + 1, (nl + nr) / 2 + 1, nr, ql, qr);
  return {b.first + max(0ll, a.first - b.second),
          a.second + max(0ll, b.second - a.first)};
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int i, j, k, l, n;
  cin >> s;
  n = s.length();
  build(1, 1, n);
  long long int q;
  cin >> q;
  while (q--) {
    cin >> k >> l;
    pair<long long int, long long int> an = query(1, 1, n, k, l);
    cout << (-an.second - an.first + l - k + 1) << "\n";
  }
  return 0;
}
