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;
vector<int> open, close, done;
string s;
void build(int x, int l, int r) {
  if (l > 0 && r <= s.size()) {
    if (l == r) {
      if (s[l - 1] == '(') {
        open[x] = 1;
        close[x] = 0;
        done[x] = 0;
      } else if (s[l - 1] == ')') {
        close[x] = 1;
        open[x] = 0;
        done[x] = 0;
      }
      return;
    } else {
      int m = (l + r) / 2;
      build(2 * x, l, m);
      build(2 * x + 1, m + 1, r);
      int yo = min(open[2 * x], close[2 * x + 1]);
      open[x] = open[2 * x] + open[2 * x + 1] - yo;
      close[x] = close[2 * x] + close[2 * x + 1] - yo;
      done[x] = done[2 * x] + done[2 * x + 1] + 2 * yo;
    }
  }
}
tuple<int, int, int> query(int x, int start, int end, int l, int r) {
  if (r < start || end < l) return make_tuple(0, 0, 0);
  if (l <= start && r >= end) return make_tuple(done[x], open[x], close[x]);
  int mid = (start + end) / 2;
  tuple<int, int, int> p1 = query(2 * x, start, mid, l, r);
  tuple<int, int, int> p2 = query(2 * x + 1, mid + 1, end, l, r);
  int yo = min(get<1>(p1), get<2>(p2));
  int d = get<0>(p1) + get<0>(p2) + 2 * yo;
  int o = get<1>(p1) + get<1>(p2) - yo;
  int c = get<2>(p1) + get<2>(p2) - yo;
  return make_tuple(d, o, c);
}
int main() {
  cin >> s;
  int m;
  cin >> m;
  int n;
  for (n = 0; s[n]; n++) {
  }
  for (int i = 0; i <= 2 * n; i++) {
    open.push_back(0);
    close.push_back(0);
    done.push_back(0);
  }
  build(1, 1, n);
  int l, r;
  for (int i = 0; i < m; i++) {
    cin >> l >> r;
    cout << get<0>(query(1, 1, n, l, r)) << "\n";
  }
}
