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;
stringstream ss;
long long mod = 1000000007LL;
int t[1 << 20], unMatch[1 << 20];
vector<int> tree[1 << 22];
void make_tree(int now, int start, int end) {
  if (start == end) {
    tree[now].push_back(t[start]);
    return;
  }
  int mid = (start + end) / 2;
  make_tree(now * 2, start, mid);
  make_tree(now * 2 + 1, mid + 1, end);
  int i = 0, j = 0, u1 = tree[now * 2].size(), u2 = tree[now * 2 + 1].size();
  while (i != u1 || j != u2) {
    if (i == u1)
      tree[now].push_back(tree[now * 2 + 1][j++]);
    else if (j == u2)
      tree[now].push_back(tree[now * 2][i++]);
    else if (tree[now * 2][i] < tree[now * 2 + 1][j])
      tree[now].push_back(tree[now * 2][i++]);
    else
      tree[now].push_back(tree[now * 2 + 1][j++]);
  }
}
int solve(int now, int start, int end, int l, int r, int low) {
  if (start == l && end == r) {
    int ret = lower_bound(tree[now].begin(), tree[now].end(), low) -
              tree[now].begin();
    int un = unMatch[end] - (start ? unMatch[start - 1] : 0);
    return tree[now].size() - ret - un;
  }
  int mid = (start + end) / 2, ret = 0;
  if (l <= mid) ret += solve(now * 2, start, mid, l, min(r, mid), low);
  if (r > mid) ret += solve(now * 2 + 1, mid + 1, end, max(l, mid + 1), r, low);
  return ret;
}
int main() {
  int n, q, l, r;
  string s;
  cin >> s;
  n = s.size();
  stack<int> st;
  fill(t, t + n, 10000000);
  for (int i = 0; i < s.size(); i++)
    if (s[i] == '(')
      st.push(i);
    else if (!st.empty())
      t[i] = st.top(), st.pop();
  unMatch[0] = (t[0] == 10000000);
  for (int i = 1; i < n; i++) unMatch[i] = unMatch[i - 1] + (t[i] == 10000000);
  make_tree(1, 0, n - 1);
  cin >> q;
  for (int i = 0; i < q; i++) {
    cin >> l >> r;
    cout << solve(1, 0, n - 1, l - 1, r - 1, l - 1) * 2 << endl;
  }
  return 0;
}
