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[4 * (int)1e6 + 2];
int open[4 * (int)1e6 + 2];
int close[4 * (int)1e6 + 2];
void build(string &second, int node, int start, int end) {
  if (start == end) {
    if (second[start] == ')')
      close[node - 1] = 1;
    else
      open[node - 1] = 1;
  } else {
    int mid = (start + end) / 2;
    build(second, 2 * node, start, mid);
    build(second, 2 * node + 1, mid + 1, end);
    int t = min(open[2 * node - 1], close[2 * node]);
    tree[node - 1] = tree[2 * node - 1] + tree[2 * node] + 2 * t;
    open[node - 1] = open[2 * node - 1] + open[2 * node] - t;
    close[node - 1] = close[2 * node - 1] + close[2 * node] - t;
  }
}
pair<int, pair<int, int>> query(int node, int l, int r, int start, int end) {
  if (start > r || end < l) {
    return {0, {0, 0}};
  }
  if (start >= l && end <= r) {
    return {tree[node - 1], {open[node - 1], close[node - 1]}};
  }
  int mid = (start >> 1) + (end >> 1) + ((start & 1) & (end & 1));
  pair<int, pair<int, int>> node1, node2;
  node1 = query(2 * node, l, r, start, mid);
  node2 = query(2 * node + 1, l, r, mid + 1, end);
  int t = min(node1.second.first, node2.second.second);
  int a = node1.first + node2.first + 2 * t;
  int b = node1.second.first + node2.second.first - t;
  int c = node1.second.second + node2.second.second - t;
  return {a, {b, c}};
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  string second;
  cin >> second;
  build(second, 1, 0, (second.size() - 1));
  int m;
  cin >> m;
  while (m--) {
    int l, r;
    cin >> l >> r;
    l--;
    r--;
    cout << query(1, l, r, 0, second.size() - 1).first << "\n";
  }
  return 0;
}
