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 MAXN = 1e6 + 10;
struct node {
  int ans, open, close;
  node() { ans = open = close = 0; }
} segTree[4 * MAXN];
int N;
string str;
void buildTree(int at, int l, int r) {
  if (l > r) return;
  if (l == r) {
    if (str[l] == '(')
      segTree[at].open = 1;
    else
      segTree[at].close = 1;
    return;
  }
  int m = (l + r) / 2;
  buildTree(2 * at, l, m);
  buildTree(2 * at + 1, m + 1, r);
  int newAdd = min(segTree[2 * at].open, segTree[2 * at + 1].close);
  segTree[at].ans = segTree[2 * at].ans + segTree[2 * at + 1].ans + newAdd;
  segTree[at].open = segTree[2 * at].open + segTree[2 * at + 1].open - newAdd;
  segTree[at].close =
      segTree[2 * at].close + segTree[2 * at + 1].close - newAdd;
}
node* query(int at, int l, int r, int ql, int qr) {
  node* ret = new node();
  if (ql > r || qr < l) return ret;
  if (ql <= l && qr >= r) {
    ret->ans = segTree[at].ans;
    ret->open = segTree[at].open;
    ret->close = segTree[at].close;
    return ret;
  }
  int m = (l + r) / 2;
  node* r1 = query(2 * at, l, m, ql, qr);
  node* r2 = query(2 * at + 1, m + 1, r, ql, qr);
  int newAdd = min(r1->open, r2->close);
  ret->ans = r1->ans + r2->ans + newAdd;
  ret->open = r1->open + r2->open - newAdd;
  ret->close = r1->close + r2->close - newAdd;
  return ret;
}
void input() {
  getline(cin, str);
  N = (int)str.size();
  str = " " + str;
  buildTree(1, 1, N);
}
void solve() {
  int Q;
  cin >> Q;
  int from, to;
  for (int i = 1; i <= Q; i++) {
    cin >> from >> to;
    cout << 2 * query(1, 1, N, from, to)->ans << "\n";
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  input();
  solve();
  return 0;
}
