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;
struct Data {
  long long a;
  long long b;
  long long c;
};
Data family[4 * 1000000];
long long player[1000000];
Data combine(Data A, Data B) {
  Data temp;
  long long t = min(A.b, B.c);
  temp.a = A.a + B.a + t;
  temp.b = A.b + B.b - t;
  temp.c = A.c + B.c - t;
  return temp;
}
void build(long long l, long long r, long long v) {
  if (l == r) {
    Data temp;
    temp.a = 0;
    temp.b = 0;
    temp.c = 0;
    if (player[l] == 1)
      temp.b = 1;
    else
      temp.c = 1;
    family[v] = temp;
  } else {
    long long m = (l + r) / 2;
    build(m + 1, r, 2 * v + 1);
    build(l, m, 2 * v);
    family[v] = combine(family[2 * v], family[2 * v + 1]);
  }
}
Data query(long long tl, long long tr, long long v, long long l, long long r) {
  if (l > r) {
    Data temp;
    temp.a = 0;
    temp.b = 0;
    temp.c = 0;
    return temp;
  }
  if (tl == l && tr == r) {
    return family[v];
  }
  long long tm = (tr + tl) / 2;
  return combine(query(tl, tm, 2 * v, l, min(tm, r)),
                 query(tm + 1, tr, 2 * v + 1, max(l, tm + 1), r));
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  string A;
  long long n;
  cin >> A >> n;
  long long m = A.length();
  for (long long i = 0; i < m; i++) {
    if (A[i] == '(')
      player[i] = 1;
    else
      player[i] = 0;
  }
  build(0, m - 1, 1);
  long long a, b;
  for (long long i = 0; i < n; i++) {
    cin >> a >> b;
    a--;
    b--;
    cout << 2 * query(0, m - 1, 1, a, b).a << "\n";
  }
  return 0;
}
