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;
int n, m, q1, q2, sag, sol;
char A[1000010];
pair<pair<int, int>, int> D[1000005 * 4];
pair<pair<int, int>, int> find(int k, int a, int b) {
  if (q1 > b || q2 < a)
    return pair<pair<int, int>, int>(pair<int, int>(0, 0), 0);
  if (q1 <= a && b <= q2) return D[k];
  pair<pair<int, int>, int> s1, s2;
  s1 = find(2 * k, a, (a + b) / 2);
  s2 = find(2 * k + 1, (a + b) / 2 + 1, b);
  return pair<pair<int, int>, int>(
      pair<int, int>(
          s2.first.first + max(0, s1.first.first - s2.first.second),
          s1.first.second + (-1 * min(0, s1.first.first - s2.first.second))),
      s1.second + s2.second + 2 * min(s1.first.first, s2.first.second));
}
pair<pair<int, int>, int> olustur(int k, int a, int b) {
  if (a == b && A[a] == '(')
    return D[k] = pair<pair<int, int>, int>(pair<int, int>(1, 0), 0);
  if (a == b && A[a] == ')')
    return D[k] = pair<pair<int, int>, int>(pair<int, int>(0, 1), 0);
  pair<pair<int, int>, int> s1 = olustur(2 * k, a, (a + b) / 2),
                            s2 = olustur(2 * k + 1, (a + b) / 2 + 1, b);
  return D[k] = pair<pair<int, int>, int>(
             pair<int, int>(
                 s2.first.first + max(0, s1.first.first - s2.first.second),
                 s1.first.second +
                     (-1 * min(0, s1.first.first - s2.first.second))),
             s1.second + s2.second + 2 * min(s1.first.first, s2.first.second));
}
int main() {
  scanf("%s", A + 1);
  n = (int)strlen(A + 1);
  olustur(1, 1, n);
  scanf("%d", &m);
  for (int j = 1; j <= m; j++) {
    sag = 0;
    sol = 0;
    scanf("%d %d", &q1, &q2);
    printf("%d\n", find(1, 1, n).second);
  }
}
