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;
int tree[4000009][3];
int a[1000009];
void init(int node, int b, int e) {
  if (b == e) {
    tree[node][a[b]] = 1;
    tree[node][1 - a[b]] = 0;
    tree[node][2] = 0;
    return;
  }
  int mid = (b + e) / 2;
  int left = node * 2;
  int right = node * 2 + 1;
  init(left, b, mid);
  init(right, mid + 1, e);
  tree[node][2] = min(tree[left][0], tree[right][1]);
  tree[node][0] = tree[left][0] - tree[node][2] + tree[right][0];
  tree[node][1] = tree[left][1] + tree[right][1] - tree[node][2];
  tree[node][2] += tree[left][2] + tree[right][2];
}
pair<int, pair<int, int>> query(int node, int b, int e, int l, int r) {
  if (l > e || r < b) return {0, {0, 0}};
  if (b >= l && e <= r) return {tree[node][2], {tree[node][0], tree[node][1]}};
  int mid = b + e >> 1;
  int left = node * 2;
  int right = node * 2 + 1;
  pair<int, pair<int, int>> p1 = query(left, b, mid, l, r);
  pair<int, pair<int, int>> p2 = query(right, mid + 1, e, l, r);
  int mn = min(p1.second.first, p2.second.second);
  int tot = p1.first + p2.first + mn;
  int lf = p1.second.first + p2.second.first - mn;
  int rf = p1.second.second + p2.second.second - mn;
  return {tot, {lf, rf}};
}
int main() {
  string s;
  cin >> s;
  int n = s.length();
  for (int i = 0; i < n; i++) {
    if (s[i] == '(')
      a[i + 1] = 0;
    else
      a[i + 1] = 1;
  }
  int q;
  cin >> q;
  init(1, 1, n);
  while (q--) {
    int l, r;
    cin >> l >> r;
    cout << 2 * query(1, 1, n, l, r).first << endl;
  }
}
