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;
struct Node {
  int m, a, b;
  Node() : m(0), a(0), b(0) {}
} seg[(1 << 21) + 5];
string s;
Node upd(const Node& l, const Node& r) {
  Node x;
  x.m = min(l.a, r.b);
  x.a = l.a + r.a - x.m;
  x.b = l.b + r.b - x.m;
  x.m += l.m + r.m;
  return x;
}
void build(int n, int l, int r) {
  if (l == r - 1) {
    if (s[l] == '(')
      seg[n].a++;
    else
      seg[n].b++;
    return;
  }
  int mid = (l + r) / 2;
  build(n * 2, l, mid);
  build(n * 2 + 1, mid, r);
  seg[n] = upd(seg[n * 2], seg[n * 2 + 1]);
}
Node cal(int n, int l, int r, int s, int e) {
  if (l == s && r == e) return seg[n];
  int mid = (l + r) / 2;
  if (e <= mid) return cal(n * 2, l, mid, s, e);
  if (s >= mid) return cal(n * 2 + 1, mid, r, s, e);
  return upd(cal(n * 2, l, mid, s, mid), cal(n * 2 + 1, mid, r, mid, e));
}
int main() {
  ios::sync_with_stdio(false);
  cin >> s;
  int q;
  cin >> q;
  int n = s.size();
  build(1, 0, n);
  for (int i = 0; i < q; i++) {
    int l, r;
    cin >> l >> r;
    l--;
    cout << cal(1, 0, n, l, r).m * 2 << endl;
  }
  return 0;
}
