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 Node {
  int val;
  int op;
  int cl;
  Node(int x, int y, int z) {
    val = x;
    op = y;
    cl = z;
  }
  Node() { Node(0, 0, 0); };
};
string s;
vector<Node> ST;
Node temp;
int n, t, x, y, t1, t2;
void update(int root, int l, int r, int pos) {
  if (l > pos || r < pos) return;
  if (l == r) {
    if (s[pos - 1] == '(') ST[root].op = 1;
    if (s[pos - 1] == ')') ST[root].cl = 1;
    return;
  }
  int mid = (l + r) / 2;
  update(root * 2, l, mid, pos);
  update(root * 2 + 1, mid + 1, r, pos);
  ST[root].val = ST[root * 2].val + ST[root * 2 + 1].val +
                 min(ST[root * 2].op - ST[root * 2].val,
                     ST[root * 2 + 1].cl - ST[root * 2 + 1].val);
  ST[root].op = ST[root * 2].op + ST[root * 2 + 1].op;
  ST[root].cl = ST[root * 2].cl + ST[root * 2 + 1].cl;
}
Node get(int root, int l, int r, int u, int v) {
  if (u > r || v < l) return Node(0, 0, 0);
  if (u == l && v == r) return ST[root];
  int mid = (l + r) / 2;
  if (v <= mid)
    return get(root * 2, l, mid, u, v);
  else if (u > mid)
    return get(root * 2 + 1, mid + 1, r, u, v);
  else {
    Node n1 = get(root * 2, l, mid, u, mid),
         n2 = get(root * 2 + 1, mid + 1, r, mid + 1, v);
    int t1 = n1.val + n2.val + min(n1.op - n1.val, n2.cl - n2.val);
    return Node(t1, n1.op + n2.op, n1.cl + n2.cl);
  }
}
int main() {
  cin >> s;
  n = s.size();
  for (int i = 0; i < 4 * n; ++i) ST.push_back(Node(0, 0, 0));
  for (int i = 1; i <= n; ++i) update(1, 1, n, i);
  cin >> t;
  while (t--) {
    cin >> x >> y;
    cout << get(1, 1, n, x, y).val * 2 << endl;
  }
  return 0;
}
