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 mxq = 100000;
const int mxn = 3000000;
string S;
int Q, N;
int L[mxq], R[mxq];
int O[mxn], C[mxn], A[mxn];
void build(int idx, int l, int r) {
  if (r - l < 2) {
    if (S[l] == '(') {
      O[idx] = 1;
    } else {
      C[idx] = 1;
    }
    return;
  }
  int mid = (l + r) >> 1;
  build(idx << 1, l, mid);
  build(idx << 1 | 1, mid, r);
  int tmp = min(O[idx << 1], C[idx << 1 | 1]);
  A[idx] = A[idx << 1] + A[idx << 1 | 1] + tmp;
  O[idx] = O[idx << 1] + O[idx << 1 | 1] - tmp;
  C[idx] = C[idx << 1] + C[idx << 1 | 1] - tmp;
}
pair<int, pair<int, int> > query(int x, int y, int idx, int l, int r) {
  if (l >= y || x >= r)
    return pair<int, pair<int, int> >(0, pair<int, int>(0, 0));
  if (x <= l && r <= y)
    return pair<int, pair<int, int> >(A[idx], pair<int, int>(O[idx], C[idx]));
  int mid = (l + r) >> 1;
  pair<int, pair<int, int> > left = query(x, y, idx << 1, l, mid);
  pair<int, pair<int, int> > right = query(x, y, idx << 1 | 1, mid, r);
  int tmp = min(left.second.first, right.second.second);
  int ans = left.first + right.first + tmp;
  int opn = left.second.first + right.second.first - tmp;
  int cls = left.second.second + right.second.second - tmp;
  return pair<int, pair<int, int> >(ans, pair<int, int>(opn, cls));
}
int main() {
  cin >> S;
  cin >> Q;
  for (int q = 0; q < Q; q++) {
    cin >> L[q] >> R[q];
    L[q]--;
  }
  N = S.size();
  for (int i = 0; i < N << 1; i++) {
    A[i] = 0;
    O[i] = 0;
    C[i] = 0;
  }
  build(1, 0, N);
  for (int q = 0; q < Q; q++) {
    pair<int, pair<int, int> > res = query(L[q], R[q], 1, 0, N);
    cout << 2 * res.first << '\n';
  }
}
