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;
template <class T>
T sqr(T x) {
  return x * x;
}
const int maxn = 1000001;
pair<int, pair<int, int> > tree[maxn * 6];
int n, m, x, y, l, r;
char str[maxn];
inline pair<int, pair<int, int> > merge(pair<int, pair<int, int> > &x,
                                        pair<int, pair<int, int> > &y) {
  int s = min(x.second.first, y.second.second);
  pair<int, pair<int, int> > ret =
      pair<int, pair<int, int> >(0, pair<int, int>(0, 0));
  ret.first = x.first + y.first + s;
  ret.second.first = x.second.first + y.second.first - s;
  ret.second.second = x.second.second + y.second.second - s;
  return ret;
}
void build(int k, int l, int r) {
  int mid = (l + r) >> 1;
  if (l == r) {
    tree[k] = pair<int, pair<int, int> >(
        0, pair<int, int>(str[l - 1] == '(', str[l - 1] == ')'));
    return;
  }
  build(k << 1, l, mid);
  build(k << 1 | 1, mid + 1, r);
  tree[k] = merge(tree[k << 1], tree[k << 1 | 1]);
}
pair<int, pair<int, int> > query(int k, int l, int r, int ll, int rr) {
  if (ll == l && rr == r) return tree[k];
  int mid = (l + r) >> 1;
  if (rr <= mid)
    return query(k << 1, l, mid, ll, rr);
  else if (ll > mid)
    return query(k << 1 | 1, mid + 1, r, ll, rr);
  else {
    pair<int, pair<int, int> > xx = query(k << 1, l, mid, ll, mid);
    pair<int, pair<int, int> > yy = query(k << 1 | 1, mid + 1, r, mid + 1, rr);
    return merge(xx, yy);
  }
}
int main() {
  scanf("%s", str);
  int n = strlen(str);
  build(1, 1, n);
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &l, &r);
    auto res = query(1, 1, n, l, r);
    printf("%d\n", res.first * 2);
  }
}
