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 = 1e6 + 10;
const long long mod = 1e9 + 7;
const long long INF = 1e18 + 10;
long long n, i, l, r, Q;
struct data_ {
  long long ans;
  long long open;
  long long close;
} emp;
data_ t[N * 4];
char a[N];
data_ merge_(data_ x, data_ y) {
  long long k = min(x.open, y.close);
  x.open -= k;
  y.close -= k;
  data_ z;
  z.ans = x.ans + y.ans + 2 * k;
  z.close = x.close + y.close;
  z.open = x.open + y.open;
  return z;
}
void build(long long ver = 1, long long tl = 1, long long tr = n) {
  if (tl == tr) {
    if (a[tl] == '(')
      t[ver].open = 1;
    else
      t[ver].close = 1;
  } else {
    long long c = (tl + tr) / 2;
    build(ver * 2, tl, c);
    build(ver * 2 + 1, c + 1, tr);
    t[ver] = merge_(t[ver * 2], t[ver * 2 + 1]);
  }
}
data_ get(long long l, long long r, long long ver = 1, long long tl = 1,
          long long tr = n) {
  if (l > r) return emp;
  if (tl == l && tr == r) return t[ver];
  long long c = (tl + tr) / 2;
  return merge_(get(l, min(r, c), ver * 2, tl, c),
                get(max(l, c + 1), r, ver * 2 + 1, c + 1, tr));
}
int main() {
  string s;
  cin >> s;
  n = s.size();
  for (i = 0; i < s.size(); ++i) a[i + 1] = s[i];
  build();
  cin >> Q;
  while (Q--) {
    cin >> l >> r;
    data_ x = get(l, r);
    cout << x.ans << "\n";
  }
}
