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, Q;
const int N = 1000002;
char str[N];
pair<pair<int, int>, int> seg[4 * N];
pair<pair<int, int>, int> combine(pair<pair<int, int>, int> c1,
                                  pair<pair<int, int>, int> c2) {
  pair<pair<int, int>, int> ret;
  int can_make = min(c1.first.first, c2.first.second);
  ret.first.first = c1.first.first - can_make + c2.first.first;
  ret.first.second = c2.first.second - can_make + c1.first.second;
  ret.second = c1.second + c2.second + can_make;
  return ret;
}
void build(int indx, int l, int r) {
  if (l == r) {
    seg[indx].first.first = str[l] == '(';
    seg[indx].first.second = str[l] == ')';
    seg[indx].second = 0;
  } else {
    build(indx * 2, l, (l + r) / 2);
    build(indx * 2 + 1, (l + r) / 2 + 1, r);
    seg[indx] = combine(seg[indx * 2], seg[indx * 2 + 1]);
  }
}
pair<pair<int, int>, int> answer(int indx, int l, int r, int i, int j) {
  if (l > j || r < i) return {{0, 0}, 0};
  if (i <= l && j >= r) {
    return seg[indx];
  }
  return combine(answer(indx * 2, l, (l + r) / 2, i, j),
                 answer(indx * 2 + 1, (l + r) / 2 + 1, r, i, j));
}
int main() {
  scanf("%s %d", str, &Q);
  n = strlen(str);
  build(1, 0, n - 1);
  while (Q--) {
    int l, r;
    scanf("%d %d", &l, &r);
    l--, r--;
    printf("%d\n", 2 * answer(1, 0, n - 1, l, r).second);
  }
  return 0;
}
