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 mod = 1e9 + 7;
const int MAX = 200005;
int bipow(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = (ans * 1LL * a) % mod;
    a = (a * 1LL * a) % mod;
    b >>= 1;
  }
  return ans;
}
bool isPrime(int x) {
  if (x == 1)
    return 0;
  else if (x == 2 || x == 3)
    return 1;
  if (x % 2 == 0 || x % 3 == 0) return 0;
  for (int i = 5; i * i <= x; i += 6) {
    if (x % i == 0 || x % (i + 2) == 0) return 0;
  }
  return 1;
}
class seg {
 public:
  vector<int> open, close, mxl;
  void print() {
    for (int i = 1; i < mxl.size(); i++) cout << mxl[i] << " ";
    cout << endl;
  }
  void build(string &s) {
    open.assign(4 * s.length(), 0);
    close.assign(4 * s.length(), 0);
    mxl.assign(4 * s.length(), 0);
    return construct(s, 1, (int)s.length(), 1);
  }
  void construct(string &s, int second, int se, int cur) {
    if (second == se) {
      if (s[second - 1] == '(') {
        open[cur] = 1;
        close[cur] = 0;
      }
      if (s[second - 1] == ')') {
        open[cur] = 0;
        close[cur] = 1;
      }
      return;
    }
    int mid = (second + se) / 2;
    construct(s, second, mid, 2 * cur);
    construct(s, mid + 1, se, 2 * cur + 1);
    int nw = min(open[2 * cur], close[2 * cur + 1]);
    mxl[cur] = mxl[2 * cur] + mxl[2 * cur + 1] + nw;
    open[cur] = open[2 * cur] + open[2 * cur + 1] - nw;
    close[cur] = close[2 * cur] + close[2 * cur + 1] - nw;
  }
  vector<int> query(int qs, int qe, int second, int se, int cur) {
    vector<int> result;
    if (qs > se || qe < second) {
      result = {0, 0, 0};
      return result;
    }
    if (second >= qs && se <= qe) {
      result = {mxl[cur], open[cur], close[cur]};
      return result;
    }
    int mid = (second + se) / 2;
    vector<int> left = query(qs, qe, second, mid, 2 * cur),
                right = query(qs, qe, mid + 1, se, 2 * cur + 1);
    result = {left[0] + right[0] + min(left[1], right[2]),
              left[1] + right[1] - min(left[1], right[2]),
              left[2] + right[2] - min(left[1], right[2])};
    return result;
  }
};
void solve() {
  string s;
  cin >> s;
  seg T;
  T.build(s);
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    int qs, qe;
    cin >> qs >> qe;
    int ans = T.query(qs, qe, 1, s.length(), 1)[0];
    cout << 2 * ans << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tc = 1;
  while (tc--) {
    solve();
  }
  return 0;
}
