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 + 10;
int n, q;
string s;
struct node {
  int res = 0, o = 0, c = 0;
} seg[N << 2];
void build(int id = 1, int l = 0, int r = n) {
  if (r - l == 1) {
    (s[l] == '(' ? seg[id].o = 1 : seg[id].c = 1);
    return;
  }
  int mid = (l + r) >> 1;
  build((id << 1), l, mid);
  build((id << 1) ^ 1, mid, r);
  seg[id].o = seg[id << 1].o + seg[(id << 1) ^ 1].o;
  seg[id].c = seg[id << 1].c + seg[(id << 1) ^ 1].c;
  seg[id].res = seg[id << 1].res + seg[(id << 1) ^ 1].res +
                min(seg[id << 1].o - seg[id << 1].res,
                    seg[(id << 1) ^ 1].c - seg[(id << 1) ^ 1].res);
}
node get(int x, int y, int id = 1, int l = 0, int r = n) {
  node ret;
  if (x >= r || y <= l) return ret;
  if (l >= x && r <= y) return seg[id];
  int mid = (l + r) >> 1;
  node s1 = get(x, y, (id << 1), l, mid);
  node s2 = get(x, y, (id << 1) ^ 1, mid, r);
  ret.o = s1.o + s2.o;
  ret.c = s1.c + s2.c;
  ret.res = s1.res + s2.res + min(s1.o - s1.res, s2.c - s2.res);
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  n = s.size();
  build();
  cin >> q;
  while (q--) {
    int l, r;
    cin >> l >> r;
    cout << 2 * get(--l, r).res << '\n';
  }
  return 0;
}
