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>
const int N = 1000010;
using namespace std;
struct node {
  int ans, uo, uc;
};
node operator+(node a, node b) {
  node c;
  int t = min(a.uo, b.uc);
  c.ans = a.ans + b.ans + t;
  c.uo = a.uo + b.uo - t;
  c.uc = a.uc + b.uc - t;
  return c;
}
node tree[4 * N];
int ans = 0;
void upd(int node, int l, int r, int i, int type) {
  if (i < l || i > r || l > r)
    return;
  else if (l == r && l == i) {
    tree[node].ans = 0;
    tree[node].uo = type;
    tree[node].uc = type ^ 1;
    return;
  }
  upd(2 * node, l, (l + r) / 2, i, type);
  upd(2 * node + 1, (l + r) / 2 + 1, r, i, type);
  tree[node] = tree[2 * node] + tree[2 * node + 1];
}
node solve(int nod, int l, int r, int ql, int qr) {
  node d, e;
  d.ans = 0, d.uo = 0, d.uc = 0;
  e.ans = 0, e.uo = 0, e.uc = 0;
  if (l > qr || ql > r || l > r)
    return d;
  else if (ql <= l && qr >= r) {
    return tree[nod];
  }
  d = solve(2 * nod, l, (l + r) / 2, ql, qr);
  e = solve(2 * nod + 1, (l + r) / 2 + 1, r, ql, qr);
  return (d + e);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  string s;
  cin >> s;
  int le = s.length();
  for (int i = 0; i < s.length(); i++) {
    if (s[i] == '(')
      upd(1, 0, le - 1, i, 1);
    else
      upd(1, 0, le - 1, i, 0);
  }
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    ans = 0;
    int l, r;
    cin >> l >> r;
    l--, r--;
    cout << 2 * solve(1, 0, le - 1, l, r).ans << endl;
  }
}
