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;
struct node {
  int a, b, c;
};
vector<node> tree;
node operator+(node aa, node bb) {
  node cc;
  int t = min(aa.b, bb.c);
  cc.a = aa.a + bb.a + t;
  cc.b = aa.b + bb.b - t;
  cc.c = aa.c + bb.c - t;
  return cc;
}
void upd(int i, int l, int r, int pos, bool t) {
  if (l > pos || r < pos) return;
  if (l == pos && l == r) {
    tree[i].a = 0;
    tree[i].b = t;
    tree[i].c = (t ^ 1);
    return;
  }
  int m = (l + r) / 2;
  upd(2 * i, l, m, pos, t);
  upd(2 * i + 1, m + 1, r, pos, t);
  tree[i] = (tree[2 * i] + tree[2 * i + 1]);
}
node query(int i, int l, int r, int nl, int nr) {
  node d, e;
  d.a = 0, d.b = 0, d.c = 0;
  e.a = 0, e.b = 0, e.c = 0;
  if (l > nr || r < nl) return d;
  if (l >= nl && r <= nr) {
    return tree[i];
  }
  int m = (l + r) / 2;
  d = query(2 * i, l, m, nl, nr);
  e = query(2 * i + 1, m + 1, r, nl, nr);
  return (d + e);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  string s;
  cin >> s;
  n = ((long long int)(s).size());
  tree.resize(4 * n);
  for (int i = 0; i < n; ++i) {
    upd(1, 0, n - 1, i, (s[i] == '('));
  }
  int q;
  cin >> q;
  while (q--) {
    int l, r;
    cin >> l >> r;
    --l;
    --r;
    cout << 2 * query(1, 0, n - 1, l, r).a << endl;
  }
  return 0;
}
