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 a[2000005], b[2000005], n;
int query(int l, int r) {
  int ans = INT_MAX;
  for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
    if (l & 1) ans = min(ans, b[l++]);
    if (r & 1) ans = min(ans, b[--r]);
  }
  return ans;
}
int main(int argc, char const *argv[]) {
  int q, l, r;
  string s;
  cin >> s;
  n = s.length();
  for (int i = 1; i <= n; i++) {
    a[i] = a[i - 1] + ((s[i - 1] == ')') ? 1 : 0);
    b[i + n] = b[i + n - 1] + ((s[i - 1] == '(') ? 1 : -1);
  }
  for (int i = n - 1; i > 0; i--) b[i] = min(b[i << 1], b[i << 1 | 1]);
  cin >> q;
  while (q--) {
    cin >> l >> r;
    cout << 2 * (a[r] - a[l - 1] - b[l + n - 1] + query(l - 1, r + 1)) << "\n";
  }
  return 0;
}
