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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cout << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
vector<long long int> vec1, vec2, vec3, primes;
long long int n = 0, m = 0, k = 0, n1 = 0, n2 = 0, n3 = 0, stt = 0, ent = 0,
              q = 0, qry = 0;
string s;
struct Node {
  long long int open, closed, valid;
  Node() {
    open = 0;
    closed = 0;
    valid = 0;
  }
};
vector<Node> tree;
void build(long long int start, long long int end, long long int node) {
  if (start == end) {
    if (s[start] == '(') {
      tree[node].open = 1;
    } else {
      tree[node].closed = 1;
    }
    return;
  }
  long long int mid = (start + end) / 2;
  build(start, mid, 2 * node + 1);
  build(mid + 1, end, 2 * node + 2);
  Node n1 = tree[2 * node + 1];
  Node n2 = tree[2 * node + 2];
  long long int t = min(n1.open, n2.closed);
  tree[node].valid = n1.valid + n2.valid + t;
  tree[node].open = n1.open + n2.open - t;
  tree[node].closed = n1.closed + n2.closed - t;
}
Node query(long long int start, long long int end, long long int node,
           long long int l, long long int r) {
  Node n3;
  if (start > end || end < l || start > r) return n3;
  if (start >= l && end <= r) {
    return tree[node];
  }
  long long int mid = (start + end) / 2;
  Node n1 = query(start, mid, 2 * node + 1, l, r);
  Node n2 = query(mid + 1, end, 2 * node + 2, l, r);
  long long int t = min(n1.open, n2.closed);
  n3.valid = n1.valid + n2.valid + t;
  n3.open = n1.open + n2.open - t;
  n3.closed = n1.closed + n2.closed - t;
  return n3;
}
void Muneer() {
  long long int ans = 0, cnt = 0, x = 0, y = 0, z = 0, a = 0, b = 0, c = 0;
  cin >> s;
  tree.resize(s.size() * 4);
  build(0, s.size() - 1, 0);
  cin >> n;
  for (long long int i = 0; i < n; i++) {
    cin >> n1 >> n2;
    cout << 2 * query(0, s.size() - 1, 0, n1 - 1, n2 - 1).valid << endl;
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t = 1;
  while (t--) Muneer();
  return 0;
}
