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;
ifstream fin("input");
void init() {
  cin.tie(NULL);
  ios_base::sync_with_stdio(false);
}
const int mx = 1e6 + 100;
string s;
int n, dp[mx];
vector<int> tree[3 * mx];
void build_tree(int index, int l, int r) {
  if (l == r) {
    tree[index].push_back(dp[l]);
  } else {
    int mid = (l + r) / 2;
    build_tree(index * 2, l, mid);
    build_tree(index * 2 + 1, mid + 1, r);
    vector<int>& left = tree[index * 2];
    vector<int>& right = tree[index * 2 + 1];
    tree[index].resize(left.size() + right.size());
    merge(left.begin(), left.end(), right.begin(), right.end(),
          tree[index].begin());
  }
}
int query(int index, int l, int r, int a, int b, int k) {
  if (r < a || l > b) {
    return 0;
  }
  if (l >= a && r <= b) {
    vector<int>& here = tree[index];
    return tree[index].size() -
           (lower_bound(here.begin(), here.end(), k) - here.begin());
  }
  int mid = (l + r) / 2;
  return query(index * 2, l, mid, a, b, k) +
         query(index * 2 + 1, mid + 1, r, a, b, k);
}
void read() {
  cin >> s;
  n = s.length();
}
void preprocess() {
  stack<int> open;
  for (int i = 1; i <= n; i++) {
    if (s[i - 1] == '(') {
      open.push(i);
    } else {
      if (open.empty()) {
        dp[i] = 0;
      } else {
        dp[i] = open.top();
        open.pop();
      }
    }
  }
}
int ans(int l, int r) { return query(1, 1, n, l, r, l) * 2; }
void solve() {
  preprocess();
  build_tree(1, 1, n);
  int t, l, r;
  cin >> t;
  while (t--) {
    cin >> l >> r;
    cout << ans(l, r) << '\n';
  }
}
int main() {
  init();
  read();
  solve();
  return 0;
}
