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 N = 1e6 + 5, M = 1e1, OO = 1000000007;
const double EPS = 0.00001;
int n, q, a, b;
string s;
struct data {
  int bal, op, cl;
  data() {}
  data(int b, int o, int c) : bal(b), op(o), cl(c) {}
};
data st[N << 2];
data comb(data a, data b) {
  data ret;
  int matched = min(a.op, b.cl);
  ret = data(a.bal + b.bal + matched, a.op + b.op - matched,
             a.cl + b.cl - matched);
  return ret;
}
void build(int p = 1, int l = 0, int r = n - 1) {
  if (l == r)
    st[p] = data(0, s[l] == '(', s[l] == ')');
  else {
    build((p << 1), l, ((l + r) >> 1));
    build((p << 1 | 1), ((l + r) >> 1) + 1, r);
    st[p] = comb(st[(p << 1)], st[(p << 1 | 1)]);
  }
}
data query(int p, int l, int r, int i, int j) {
  if (l > r || l > j || r < i) return data(0, 0, 0);
  if (l >= i && r <= j) return st[p];
  return comb(query((p << 1), l, ((l + r) >> 1), i, j),
              query((p << 1 | 1), ((l + r) >> 1) + 1, r, i, j));
}
int main() {
  cin.sync_with_stdio(0);
  cin >> s >> q;
  n = s.length();
  build();
  while (q--) {
    cin >> a >> b;
    printf("%d\n", 2 * query(1, 0, n - 1, a - 1, b - 1).bal);
  }
  return 0;
}
