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;
int close[4 * maxn], open[4 * maxn], cnt[4 * maxn], n, q;
vector<int> seg;
string str;
void up(int idx) {
  int L = 2 * idx + 1, R = 2 * idx + 2;
  cnt[idx] = cnt[L] + cnt[R];
  cnt[idx] += min(close[R], open[L]);
  close[idx] = (close[L] + close[R]) - min(close[R], open[L]);
  open[idx] = (open[R] + open[L]) - min(close[R], open[L]);
}
void make(int s, int e, int id, char c, int idx) {
  if (s == e - 1) {
    if (c == ')') close[idx] += 1;
    if (c == '(') open[idx] += 1;
    return;
  }
  int mid = (s + e) / 2, L = 2 * idx + 1, R = 2 * idx + 2;
  if (id < mid) make(s, mid, id, c, L);
  if (id >= mid) make(mid, e, id, c, R);
  up(idx);
}
void gets(int s, int e, int l, int r, int idx) {
  if (e <= l || r <= s) return;
  if (l == s and e == r) {
    seg.push_back(idx);
    return;
  }
  int mid = (s + e) / 2, L = 2 * idx + 1, R = 2 * idx + 2;
  gets(s, mid, l, min(mid, r), L);
  gets(mid, e, max(l, mid), r, R);
}
int ans() {
  int ans = 0, o = 0, c = 0;
  for (int i = 0; i < seg.size(); i++) {
    ans += cnt[seg[i]];
    o += open[seg[i]];
    if (i + 1 < seg.size()) {
      ans += min(o, close[seg[i + 1]]);
      o -= min(o, close[seg[i + 1]]);
    }
  }
  seg.clear();
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> str;
  n = str.size();
  for (int i = 0; i < n; i++) make(0, n, i, str[i], 0);
  cin >> q;
  while (q--) {
    int r, l;
    cin >> l >> r;
    l--;
    gets(0, n, l, r, 0);
    cout << 2 * ans() << endl;
  }
  return 0;
}
