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 N = 1000001;
long long A[N];
long long st[N << 2];
pair<pair<long long, long long>, long long> par[N];
long long rep[N];
void update(long long pos, long long val, long long l, long long r,
            long long nod) {
  if (pos < l || r < pos) return;
  if (pos <= l && r <= pos) {
    st[nod] = val;
    return;
  }
  long long med = (l + r) / 2;
  update(pos, val, l, med, nod * 2);
  update(pos, val, med + 1, r, nod * 2 + 1);
  st[nod] = st[nod * 2] + st[nod * 2 + 1];
  return;
}
long long query(long long ini, long long fin, long long l, long long r,
                long long nod) {
  if (fin < l || r < ini) return 0;
  if (ini <= l && r <= fin) return st[nod];
  long long med = (l + r) / 2;
  long long L = query(ini, fin, l, med, nod * 2);
  long long R = query(ini, fin, med + 1, r, nod * 2 + 1);
  return L + R;
}
int main() {
  char k[N];
  scanf("%s", k);
  long long n = strlen(k);
  long long Q, l, r;
  scanf("%lld", &Q);
  for (long long i = 0; i < Q; i++) {
    scanf("%lld %lld", &l, &r);
    par[i] = make_pair(make_pair(r - 1, l - 1), i);
  }
  sort(par, par + Q);
  deque<pair<long long, long long> > q;
  long long pos = 0;
  for (long long i = 0; i < n; i++) {
    if (q.empty()) {
      q.push_back(make_pair(k[i], i));
    } else {
      pair<long long, long long> ac = q.back();
      if (char(ac.first) == '(' && k[i] == ')') {
        update(ac.second, 1, 0, n - 1, 1);
        q.pop_back();
      } else {
        q.push_back(make_pair(k[i], i));
      }
    }
    while (pos < Q && par[pos].first.first == i) {
      rep[par[pos].second] = query(par[pos].first.second, i, 0, n - 1, 1);
      pos++;
    }
  }
  for (long long i = 0; i < Q; i++) {
    printf("%lld\n", rep[i] * 2);
  }
  return 0;
}
