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 + 6;
int INF = 1e9;
int a[4 * N], b[4 * N], c[4 * N];
string s;
struct Node {
  int a, b, c;
  Node(int a, int b, int c) : a(a), b(b), c(c) {}
};
void build(int p, int l, int r) {
  if (l == r) {
    a[p] = 0;
    b[p] = s[l] == '(';
    c[p] = s[l] == ')';
    return;
  }
  build(2 * p, l, (l + r) / 2);
  build(2 * p + 1, (l + r) / 2 + 1, r);
  int p1, p2;
  p1 = 2 * p, p2 = 2 * p + 1;
  int t = min(b[p1], c[p2]);
  a[p] = a[p1] + a[p2] + t;
  b[p] = b[p1] + b[p2] - t;
  c[p] = c[p1] + c[p2] - t;
}
Node query(int p, int l, int r, int i, int j) {
  if (i > r || j < l) return Node(0, 0, 0);
  if (l >= i && r <= j) return Node(a[p], b[p], c[p]);
  Node p1 = query(2 * p, l, (l + r) / 2, i, j);
  Node p2 = query(2 * p + 1, (l + r) / 2 + 1, r, i, j);
  Node q(0, 0, 0);
  int t = min(p1.b, p2.c);
  q.a = p1.a + p2.a + t;
  q.b = p1.b + p2.b - t;
  q.c = p1.c + p2.c - t;
  return q;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> s >> m;
  n = s.size();
  build(1, 0, n - 1);
  for (int i = 0; i < m; i++) {
    int l, r;
    cin >> l >> r;
    l--, r--;
    cout << 2 * query(1, 0, n - 1, l, r).a << "\n";
  }
}
