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;
const int N = 1e6 + 5;
string s;
int n;
struct Node {
  int cnt = 0;
  int open = 0;
  int close = 0;
  Node() { cnt = open = close = 0; }
};
Node tree[2 * N];
Node temp;
Node merge(Node x, Node y) {
  int v = min(x.open, y.close);
  temp.cnt = x.cnt + y.cnt + (2 * v);
  temp.open = x.open + y.open - v;
  temp.close = x.close + y.close - v;
  return temp;
}
void build(string s) {
  for (int i = 0; i < n; ++i) {
    tree[i + n].cnt = 0;
    tree[i + n].open = (s[i] == '(');
    tree[i + n].close = (s[i] == ')');
  }
  for (int i = n - 1; i > 0; i--) {
    tree[i] = merge(tree[i << 1], tree[i << 1 | 1]);
  }
}
int query(int l, int r) {
  Node al, ar;
  for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
    if (l & 1) {
      al = merge(al, tree[l++]);
    }
    if (r & 1) {
      ar = merge(tree[--r], ar);
    }
  }
  Node a = merge(al, ar);
  return a.cnt;
}
int main() {
  cin >> s;
  n = s.length();
  build(s);
  int t;
  cin >> t;
  while (t--) {
    int l, r;
    cin >> l >> r;
    l--;
    r--;
    cout << query(l, r + 1) << endl;
  }
}
