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;
const int maxn = 1000000;
string s;
int t[4 * (maxn + 100) + 100];
int o[4 * (maxn + 100) + 100];
int c[4 * (maxn + 100) + 100];
void build(int v, int tl, int tr) {
  if (tl == tr) {
    if (s[tl] == '(') {
      o[v] = 1;
    } else {
      c[v] = 1;
    }
  } else {
    int tm = tl + (tr - tl) / 2;
    build(2 * v, tl, tm);
    build(2 * v + 1, tm + 1, tr);
    int newt = min(o[2 * v], c[2 * v + 1]);
    t[v] = t[2 * v] + t[2 * v + 1] + newt;
    o[v] = o[2 * v] + o[2 * v + 1] - newt;
    c[v] = c[2 * v] + c[2 * v + 1] - newt;
  }
}
tuple<int, int, int> query(int v, int tl, int tr, int l, int r) {
  if (l > r) {
    return make_tuple(0, 0, 0);
  }
  if (tl == l && tr == r) {
    return make_tuple(t[v], o[v], c[v]);
  }
  int tm = tl + (tr - tl) / 2;
  tuple<int, int, int> a = query(2 * v, tl, tm, l, min(r, tm)),
                       b = query(2 * v + 1, tm + 1, tr, max(l, tm + 1), r);
  int newt = min(get<1>(a), get<2>(b));
  int T = get<0>(a) + get<0>(b) + newt;
  int O = get<1>(a) + get<1>(b) - newt;
  int C = get<2>(a) + get<2>(b) - newt;
  return make_tuple(T, O, C);
}
int main(void) {
  cin >> s;
  int n = s.size();
  build(1, 0, n - 1);
  int m;
  cin >> m;
  int il, ir;
  for (int i = 0; i < m; i++) {
    cin >> il >> ir;
    cout << 2 * get<0>(query(1, 0, n - 1, il - 1, ir - 1)) << "\n";
  }
  return 0;
}
