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;
pair<int, int> seg_tree[4000000];
string s;
void build(int start, int end, int node) {
  if (start == end) {
    seg_tree[node] = {0, 0};
    if (s[start] == ')')
      seg_tree[node].first = 1;
    else
      seg_tree[node].second = 1;
  } else {
    int mid = (start + end) / 2;
    build(start, mid, 2 * node);
    build(mid + 1, end, 2 * node + 1);
    seg_tree[node].first = seg_tree[2 * node].first;
    seg_tree[node].second = seg_tree[2 * node + 1].second;
    if (seg_tree[2 * node].second > seg_tree[2 * node + 1].first) {
      seg_tree[node].second +=
          (seg_tree[2 * node].second - seg_tree[2 * node + 1].first);
    } else
      seg_tree[node].first +=
          (seg_tree[2 * node + 1].first - seg_tree[2 * node].second);
  }
}
pair<int, int> query(int start, int end, int node, int l, int r) {
  if (l == start && r == end) return seg_tree[node];
  int mid = (start + end) / 2;
  if (r <= mid)
    return query(start, mid, 2 * node, l, r);
  else if (l > mid)
    return query(mid + 1, end, 2 * node + 1, l, r);
  else {
    pair<int, int> p = query(start, mid, 2 * node, l, mid);
    pair<int, int> q = query(mid + 1, end, 2 * node + 1, mid + 1, r);
    pair<int, int> x;
    x.first = p.first;
    x.second = q.second;
    if (p.second > q.first) {
      x.second += (p.second - q.first);
    } else
      x.first += (q.first - p.second);
    return x;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> s;
  int n = s.size();
  build(0, n - 1, 1);
  int m;
  cin >> m;
  while (m--) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    pair<int, int> p = query(0, n - 1, 1, x, y);
    cout << (y - x + 1) - (p.first + p.second) << endl;
  }
  return 0;
}
