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;
string s;
struct node {
  int open = 0, close = 0;
};
struct node tree[1000005 << 2];
node build(int b, int e, int node) {
  struct node &t = tree[node];
  if (b == e) {
    if (s[b - 1] == '(')
      t.open = 1;
    else
      t.close = 1;
    return t;
  }
  int left = node << 1;
  int right = left + 1;
  int mid = (b + e) >> 1;
  struct node l, r;
  l = build(b, mid, left);
  r = build(mid + 1, e, right);
  int x = min(l.open, r.close);
  l.open -= x;
  r.close -= x;
  t.open = l.open + r.open;
  t.close = l.close + r.close;
  return t;
}
int start, finish;
node query(int b, int e, int node) {
  struct node t = tree[node];
  if (b > finish || e < start) {
    struct node x;
    return x;
  };
  if (b >= start && e <= finish) {
    return t;
  }
  int left = node << 1;
  int right = left + 1;
  int mid = (b + e) >> 1;
  struct node l, r;
  l = query(b, mid, left);
  r = query(mid + 1, e, right);
  int x = min(l.open, r.close);
  l.open -= x;
  r.close -= x;
  t.open = l.open + r.open;
  t.close = l.close + r.close;
  return t;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> s;
  build(1, s.size(), 1);
  int m;
  cin >> m;
  while (m--) {
    cin >> start >> finish;
    node x = query(1, s.size(), 1);
    cout << (finish - start + 1) - x.open - x.close << endl;
  }
  return 0;
}
