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;
string st;
struct str {
  int total, t_open, t_end;
};
str tree[4 * 1000007];
str combine(str a, str b) {
  int mn = min(a.t_open, b.t_end);
  return {a.total + b.total + mn, a.t_open + b.t_open - mn,
          a.t_end + b.t_end - mn};
}
void build(int node, int l, int r) {
  if (l == r) {
    tree[node] = {0, st[l - 1] == '(' ? 1 : 0, st[l - 1] == ')' ? 1 : 0};
    return;
  }
  int mid = (l + r) / 2;
  build(node * 2, l, mid);
  build(node * 2 + 1, mid + 1, r);
  tree[node] = combine(tree[node * 2], tree[node * 2 + 1]);
}
str query(int node, int l, int r, int i, int j) {
  if (l > j || r < i) return {-1, 0, 0};
  if (l >= i && r <= j) return tree[node];
  int mid = (l + r) / 2;
  str le = query(node * 2, l, mid, i, j);
  str ri = query(node * 2 + 1, mid + 1, r, i, j);
  if (le.total == -1) return ri;
  if (ri.total == -1) return le;
  return combine(le, ri);
}
int main() {
  int i, j, l, r, q;
  str ans;
  cin >> st;
  build(1, 1, st.size());
  cin >> q;
  for (i = 0; i < q; i++) {
    cin >> l >> r;
    ans = query(1, 1, st.size(), l, r);
    cout << (ans.total * 2) << "\n";
  }
  return 0;
}
