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;
void da(int *a, int s, int e) {
  for (int i = s; i < e; i++) cout << a[i] << " ";
}
void db(string s) { cout << s << '\n'; }
void cp(int *a, int *b, int n) {
  for (int i = 0; i < n; i++) b[i] = a[i];
}
void de(int a) { cout << a << " <\n"; }
const long long N = 1e6 + 10, MOD = 1e9 + 7, INF = 1e15 + 10;
int n, a[N], tmp[N], m;
string s;
struct node {
  int ob, cb, ans;
  node() {}
  node(int ob, int cb, int ans) : ob(ob), cb(cb), ans(ans) {}
} val[4 * N];
node merge(node a, node b) {
  int tmp = min(a.ob, b.cb);
  return node(a.ob + b.ob - tmp, a.cb + b.cb - tmp, a.ans + b.ans + tmp);
}
void build(int l = 0, int r = n, int id = 1) {
  if (r - l == 1) {
    val[id] = node(0, 0, 0);
    if (s[l] == '(')
      val[id].ob = 1;
    else
      val[id].cb = 1;
    return;
  }
  build(l, (l + r) >> 1, id * 2);
  build((l + r) >> 1, r, id * 2 | 1);
  val[id] = merge(val[id * 2], val[id * 2 | 1]);
}
node get(int s, int e, int l = 0, int r = n, int id = 1) {
  if (e <= l || s >= r) return node(0, 0, 0);
  if (s <= l && e >= r) return val[id];
  return merge(get(s, e, l, (l + r) >> 1, id * 2),
               get(s, e, (l + r) >> 1, r, id * 2 | 1));
}
int main() {
  cin >> s >> m;
  n = s.length();
  build();
  for (int i = 0; i < m; i++) {
    int l, r;
    cin >> l >> r;
    l--;
    node res = get(l, r);
    cout << res.ans * 2 << '\n';
  }
}
