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 maxn = 1000001;
string s;
struct segNode {
  int open;
  int close;
  int cnt;
} seg[4 * maxn + 1];
void build(int si, int ss, int se) {
  if (ss == se) {
    if (s[ss] == '(')
      seg[si].open = 1;
    else
      seg[si].close = 1;
    return;
  }
  int mid = (ss + se) / 2;
  build(2 * si, ss, mid);
  build(2 * si + 1, mid + 1, se);
  int minc = min(seg[2 * si].open, seg[2 * si + 1].close);
  seg[si].open = seg[2 * si].open + seg[2 * si + 1].open - minc;
  seg[si].close = seg[2 * si].close + seg[2 * si + 1].close - minc;
  seg[si].cnt = seg[2 * si].cnt + seg[2 * si + 1].cnt + 2 * minc;
}
segNode query(int si, int ss, int se, int qs, int qe) {
  if (qs > se || qe < ss) {
    seg[0].close = 0;
    seg[0].open = 0;
    seg[0].cnt = 0;
    return seg[0];
  }
  if (ss >= qs && se <= qe) return seg[si];
  int mid = (ss + se) / 2;
  segNode l = query(2 * si, ss, mid, qs, qe);
  segNode r = query(2 * si + 1, mid + 1, se, qs, qe);
  int minc = min(l.open, r.close);
  segNode x;
  x.open = l.open + r.open - minc;
  x.close = l.close + r.close - minc;
  x.cnt = l.cnt + r.cnt + 2 * minc;
  return x;
}
int main() {
  cin >> s;
  build(1, 0, (int)s.length() - 1);
  int m, l, r;
  cin >> m;
  while (m--) {
    cin >> l >> r;
    cout << query(1, 0, s.length() - 1, l - 1, r - 1).cnt << "\n";
  }
  return 0;
}
