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;
using vvi = vector<vector<int> >;
using vvl = vector<vector<long long> >;
const long long M = 1e6;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
long long binpow(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % M;
    a = a * a % M;
    b >>= 1;
  }
  return res;
}
template <class T>
struct bit {
  vector<T> b;
  int n;
  bit(int n) {
    this->n = n + 100;
    b.assign(n + 100, 0);
  }
  T sum(int i) {
    T ret = 0;
    for (++i; i > 0; i -= i & -i) ret += b[i];
    return ret;
  }
  T sum(int l, int r) { return sum(r) - sum(l - 1); }
  void add(int i, T x) {
    for (++i; i < n; i += i & -i) b[i] += x;
  }
};
void solve() {
  string s;
  cin >> s;
  stack<int> st;
  int n = s.size();
  int q;
  cin >> q;
  vector<tuple<int, int, int> > qr(q);
  for (int i = 0; i < q; i++) {
    int l, r;
    cin >> l >> r;
    l--;
    r--;
    qr[i] = {l, r, i};
  }
  sort(qr.begin(), qr.end(),
       [&](tuple<int, int, int> a, tuple<int, int, int> b) {
         return get<1>(a) < get<1>(b);
       });
  bit<int> tree(n);
  int ldp = -1;
  vector<int> ans(q);
  for (auto [l, r, id] : qr) {
    while (ldp < r) {
      ldp++;
      if (s[ldp] == '(') {
        st.emplace(ldp);
      } else {
        if (!st.empty()) {
          tree.add(st.top(), 2);
          st.pop();
        }
      }
    }
    ans[id] = tree.sum(l, r);
  }
  {
    for (auto _pr : ans) cout << _pr << " ";
    cout << "\n";
  };
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
}
