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;
struct node {
  int open = 0, close = 0, done = 0;
  void merge(node &a, node &b) {
    int mn = min(a.open, b.close);
    done += a.done + b.done + 2 * mn;
    open = a.open + b.open - mn;
    close = a.close + b.close - mn;
  }
};
vector<node> seg(1 << 22);
node nullnode;
int n;
void build(string &s, int l, int r, int v = 1) {
  if (l == r) {
    if (s[l] == '(')
      seg[v].open = 1;
    else
      seg[v].close = 1;
    return;
  }
  int mid = (l + r) >> 1;
  build(s, l, mid, v * 2);
  build(s, mid + 1, r, v * 2 + 1);
  seg[v].merge(seg[v * 2], seg[v * 2 + 1]);
  return;
}
node query(int l, int r, int v = 1, int start = 0, int end = n - 1) {
  if (r < start || l > end) return nullnode;
  if (l <= start && r >= end) return seg[v];
  int mid = (start + end) >> 1;
  node q1 = query(l, r, v * 2, start, mid);
  node q2 = query(l, r, v * 2 + 1, mid + 1, end);
  node ans;
  ans.merge(q1, q2);
  return ans;
}
void solve() {
  string s;
  cin >> s;
  n = s.length();
  build(s, 0, n - 1);
  int m;
  cin >> m;
  while (m--) {
    int l, r;
    cin >> l >> r;
    l--, r--;
    node ans = query(l, r);
    cout << ans.done << "\n";
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
