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;
int di[] = {0, 0, 1, -1, 1, 1, -1, -1};
int dj[] = {1, -1, 0, 0, -1, 1, 1, -1};
const double PI = 3.14159265359, EPS = 1e-9;
const int N = 1000006, M = 1000006, Mod = 1000000007;
vector<int> adj[N];
pair<int, int> tree[4 * N];
int n;
string str;
void build(int idx, int s, int e) {
  if (s == e) {
    if (str[s] == '(') ++tree[idx].first;
    if (str[s] == ')') ++tree[idx].second;
    return;
  }
  int m = (s + e) / 2;
  build(idx << 1, s, m);
  build(idx << 1 | 1, m + 1, e);
  tree[idx].first = tree[idx << 1 | 1].first +
                    max(0, tree[idx << 1].first - tree[idx << 1 | 1].second);
  tree[idx].second = tree[idx << 1].second +
                     max(0, tree[idx << 1 | 1].second - tree[idx << 1].first);
}
pair<int, int> maxx(int l, int r, int idx, int s, int e) {
  if (l > e || r < s) return pair<int, int>(0, 0);
  if (l <= s && r >= e) return tree[idx];
  int m = (s + e) / 2;
  pair<int, int> p1 = maxx(l, r, idx << 1, s, m);
  pair<int, int> p2 = maxx(l, r, idx << 1 | 1, m + 1, e);
  pair<int, int> ret;
  ret.first = p2.first + max(0, p1.first - p2.second);
  ret.second = p1.second + max(0, p2.second - p1.first);
  return ret;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0), cout.precision(10),
      cout << fixed;
  cin >> str;
  str = "s" + str;
  n = (int)str.size();
  build(1, 1, n);
  int m;
  cin >> m;
  int l, r;
  while (m--) {
    cin >> l >> r;
    pair<int, int> p = maxx(l, r, 1, 1, n);
    cout << r - l + 1 - p.first - p.second << '\n';
  }
  return 0;
}
