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;
string a;
struct info {
  int open, close, match;
  info() {}
  info(int a, int b, int c) { open = a, close = b, match = c; }
  info operator+(info &right) {
    int temp = min(open, right.close);
    return info(open + right.open - temp, close + right.close - temp,
                match + right.match + temp);
  }
};
info tree[3 * 1000001];
void init(int n, int b, int e) {
  if (b == e) {
    tree[n] = (a[b - 1] == '(') ? info(1, 0, 0) : info(0, 1, 0);
    return;
  }
  int m = (b + e) / 2;
  int left = 2 * n, right = left + 1;
  init(left, b, m);
  init(right, m + 1, e);
  tree[n] = tree[left] + tree[right];
}
info query(int n, int b, int e, int l, int r) {
  if (e < l || b > r) return info(0, 0, 0);
  if (b >= l && e <= r) return tree[n];
  int m = (b + e) / 2;
  int left = 2 * n, right = left + 1;
  info p1 = query(left, b, m, l, r);
  info p2 = query(right, m + 1, e, l, r);
  info temp = p1 + p2;
  return temp;
}
int main() {
  cin >> a;
  init(1, 1, a.size());
  int q;
  cin >> q;
  int l, r;
  while (q--) {
    cin >> l >> r;
    cout << 2 * (query(1, 1, a.size(), l, r)).match << endl;
  }
  return 0;
}
