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;
long long MOD = 1e9 + 7;
long long g, h, l;
const long long N = 1000005, inf = 1e9 + 5;
string s;
struct poi {
  long long left, right, corr;
  poi() { left = right = corr = 0; }
};
poi tree[4 * N];
void build(long long node, long long st, long long en) {
  if (st == en) {
    tree[node].left = tree[node].right = tree[node].corr = 0;
    if (s[st] == '(') {
      tree[node].left = 1;
    } else {
      tree[node].right = 1;
    }
    return;
  }
  long long mid = (st + en) / 2;
  build(node * 2, st, mid);
  build(node * 2 + 1, mid + 1, en);
  tree[node].corr = tree[node * 2].corr + tree[node * 2 + 1].corr +
                    2 * min(tree[node * 2].left, tree[node * 2 + 1].right);
  tree[node].left = tree[node * 2].left -
                    min(tree[node * 2].left, tree[node * 2 + 1].right) +
                    tree[node * 2 + 1].left;
  tree[node].right = tree[node * 2].right -
                     min(tree[node * 2].left, tree[node * 2 + 1].right) +
                     tree[node * 2 + 1].right;
}
poi query(long long node, long long st, long long en, long long l,
          long long r) {
  if (st > en or st > r or en < l) {
    poi ret;
    return ret;
  }
  if (l <= st and en <= r) {
    return tree[node];
  }
  long long mid = (st + en) / 2;
  auto q1 = query(node * 2, st, mid, l, r);
  auto q2 = query(node * 2 + 1, mid + 1, en, l, r);
  poi ret;
  ret.corr = q1.corr + q2.corr + 2 * min(q1.left, q2.right);
  ret.left = q1.left - min(q1.left, q2.right) + q2.left;
  ret.right = q1.right - min(q1.left, q2.right) + q2.right;
  return ret;
}
long long add(long long a, long long b);
long long mul(long long a, long long b);
long long binpo(long long a, long long b);
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> s;
  long long n = (long long)s.size();
  s = '#' + s;
  long long m;
  cin >> m;
  build(1, 1, n);
  for (long long i = 0; i < m; i++) {
    cin >> g >> h;
    cout << query(1, 1, n, g, h).corr << endl;
  }
}
long long add(long long a, long long b) { return (a % MOD + b % MOD) % MOD; }
long long mul(long long a, long long b) { return (a * 1ll * b) % MOD; }
long long binpo(long long a, long long b) {
  int ans = 1;
  while (b) {
    if (b % 2 != 0) {
      ans = mul(ans, a);
    }
    a = mul(a, a);
    b = b / 2;
  }
  return ans;
}
