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;
long double pi = 3.14159265358979323;
const double EPS = 1e-12;
const int N = 1e6 + 5;
const int mod = 1e9 + 7;
class node {
 public:
  int op, cls, ans;
  node() { op = 0, cls = 0, ans = 0; }
  node operator+(node &t) {
    node obj;
    int cmn = min(op, t.cls);
    obj.ans = ans + t.ans + cmn * 2;
    obj.op = op + t.op - cmn;
    obj.cls = cls + t.cls - cmn;
    return obj;
  }
};
string a;
node tree[N << 2];
void build(long long node, long long l, long long r) {
  if (l == r) {
    tree[node].op = a[l] == '(';
    tree[node].cls = a[l] == ')';
    return;
  }
  long long mid = (l + r) >> 1;
  build(node << 1, l, mid);
  build((node << 1) + 1, mid + 1, r);
  tree[node] = tree[node << 1] + tree[(node << 1) + 1];
}
node P;
node query(long long nod, long long l, long long r, long long ql,
           long long qr) {
  if (r < ql || l > qr) return P;
  if (l >= ql && r <= qr) return tree[nod];
  long long mid = (l + r) >> 1;
  node A1 = query(nod << 1, l, mid, ql, qr);
  node A2 = query((nod << 1) + 1, mid + 1, r, ql, qr);
  return A1 + A2;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> a;
  a = '#' + a;
  long long n = a.size();
  n--;
  build(1, 1, n);
  long long q;
  cin >> q;
  while (q--) {
    long long l, r;
    cin >> l >> r;
    node Ans = query(1, 1, n, l, r);
    cout << Ans.ans << "\n";
  }
  return 0;
}
