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 maxN = 4e6;
template <typename T>
T In() {
  T x;
  cin >> x;
  return x;
}
string S;
struct node {
  int L = 0, R = 0, A = 0;
} T[maxN];
void build(int id, int l, int r) {
  if (l + 1 == r) {
    if (S[l] == ')')
      T[id].R = 1;
    else
      T[id].L = 1;
    return;
  }
  int m = (l + r) >> 1;
  build((id << 1) + 0, l, m);
  build((id << 1) + 1, m, r);
  int shams = min(T[(id << 1)].L, T[(id << 1) + 1].R);
  T[id].A = T[(id << 1)].A + T[(id << 1) + 1].A + shams;
  T[id].L = T[(id << 1)].L + T[(id << 1) + 1].L - shams;
  T[id].R = T[(id << 1)].R + T[(id << 1) + 1].R - shams;
}
node get(int id, int l, int r, int s, int e) {
  if (e < s) swap(s, e);
  if (r <= s || e + 1 <= l) {
    node ans;
    ans.A = ans.L = ans.R = 0;
    return ans;
  }
  if (s <= l && r <= e + 1) return T[id];
  int m = (l + r) >> 1;
  node c1 = get((id << 1), l, m, s, e), c2 = get((id << 1) + 1, m, r, s, e),
       ans;
  int shams = min(c1.L, c2.R);
  ans.A = c1.A + c2.A + shams;
  ans.L = c1.L + c2.L - shams;
  ans.R = c1.R + c2.R - shams;
  return ans;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  S = In<string>();
  build(1, 0, S.size());
  int m = In<int>();
  for (int i = 0; i < m; i++)
    cout << 2 * get(1, 0, S.size(), In<int>() - 1, In<int>() - 1).A << "\n";
  return cout << endl, 0;
}
