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 long long N = 1000006;
struct node {
  long long op, cl, bal;
  node operator+(node a) {
    long long m = min(op, a.cl);
    return {op + a.op - m, cl + a.cl - m, bal + a.bal + 2 * m};
  }
};
node tree[4 * N];
string s;
void build(long long n, long long tl, long long tr) {
  if (tl == tr) {
    tree[n] = {s[tl] == '(', s[tl] == ')', 0};
    return;
  }
  long long tm = (tl + tr) / 2;
  build(2 * n, tl, tm);
  build(2 * n + 1, tm + 1, tr);
  tree[n] = tree[2 * n] + tree[2 * n + 1];
}
node query(long long n, long long tl, long long tr, long long l, long long r) {
  if (l > r) return {0, 0, 0};
  if (tl == l and tr == r) return tree[n];
  long long tm = (tl + tr) / 2;
  node al = query(2 * n, tl, tm, l, min(r, tm));
  node ar = query(2 * n + 1, tm + 1, tr, max(l, tm + 1), r);
  return al + ar;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  long long l, r, q, n = s.size();
  s = "#" + s;
  build(1, 1, n);
  cin >> q;
  while (q--) {
    cin >> l >> r;
    node ans = query(1, 1, n, l, r);
    cout << ans.bal << "\n";
  }
}
