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;
const int N = 1e6 + 10;
int tree[N << 2], open[N], close[N];
void build(int node, int st, int end) {
  if (st == end) {
    tree[node] = open[st];
    return;
  }
  int mid = (st + end) >> 1;
  build(node << 1, st, mid);
  build((node << 1) + 1, mid + 1, end);
  tree[node] = min(tree[node << 1], tree[(node << 1) + 1]);
}
int query(int node, int st, int end, int l, int r) {
  if (r < st || l > end) {
    return INT_MAX;
  }
  if (l <= st && end <= r) {
    return tree[node];
  }
  int mid = (st + end) >> 1;
  return (min(query(node << 1, st, mid, l, r),
              query((node << 1) + 1, mid + 1, end, l, r)));
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> s;
  int n = s.size();
  for (int i = 1; i <= n; i++) {
    if (s[i - 1] == '(') {
      open[i] = 1;
    } else {
      open[i] = -1;
      close[i] = 1;
    }
  }
  for (int i = 1; i <= n; i++) {
    open[i] += open[i - 1];
    close[i] += close[i - 1];
  }
  build(1, 1, n);
  int q;
  cin >> q;
  while (q--) {
    int l, r;
    cin >> l >> r;
    int mini = query(1, 1, n, max(1, l - 1), r);
    cout << 2 * ((close[r] - close[l - 1]) - max(0, (open[l - 1] - mini)))
         << '\n';
  }
  return 0;
}
