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;
vector<int> a, b, c;
string S;
int sz;
void buildTree(int s = 0, int e = sz, int treeNode = 1) {
  if (s == e) {
    a[treeNode] = 0;
    if (S[e] == '(') {
      b[treeNode] = 1;
    } else {
      c[treeNode] = 1;
    }
    return;
  }
  int mid = (s + e) / 2;
  buildTree(s, mid, 2 * treeNode);
  buildTree(mid + 1, e, 2 * treeNode + 1);
  int t = min(b[treeNode * 2], c[2 * treeNode + 1]);
  a[treeNode] = a[treeNode * 2] + a[2 * treeNode + 1] + t;
  b[treeNode] = b[treeNode * 2] + b[2 * treeNode + 1] - t;
  c[treeNode] = c[treeNode * 2] + c[2 * treeNode + 1] - t;
}
pair<int, pair<int, int> > query(int l, int r, int s = 0, int e = sz,
                                 int treeNode = 1) {
  if (s > r || l > e)
    return pair<int, pair<int, int> >(0, pair<int, int>(0, 0));
  if (s >= l && e <= r)
    return pair<int, pair<int, int> >(a[treeNode],
                                      pair<int, int>(b[treeNode], c[treeNode]));
  int mid = (s + e) / 2;
  pair<int, pair<int, int> > ans1 = query(l, r, s, mid, 2 * treeNode);
  pair<int, pair<int, int> > ans2 = query(l, r, mid + 1, e, 2 * treeNode + 1);
  int t = min(ans1.second.first, ans2.second.second);
  ans1.first = ans1.first + ans2.first + t;
  ans1.second.first = ans1.second.first + ans2.second.first - t;
  ans1.second.second = ans1.second.second + ans2.second.second - t;
  return ans1;
}
int main() {
  cin >> S;
  sz = S.length() - 1;
  a.resize(S.length() * 4);
  b.resize(S.length() * 4);
  c.resize(S.length() * 4);
  buildTree();
  int K;
  cin >> K;
  while (K--) {
    int l, r;
    cin >> l >> r;
    cout << 2 * (query(l - 1, r - 1).first) << endl;
  }
  return 0;
}
