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 = 1000010;
struct data {
  int ok, open, close;
  data() {}
  data(int _ok, int _open, int _close) {
    ok = _ok, open = _open, close = _close;
  }
  data operator+(const data &temp) const {
    data ret;
    ret = data(ok + min(open, temp.close) + temp.ok,
               temp.open + open - min(open, temp.close),
               close + temp.close - min(open, temp.close));
    return ret;
  }
  void print() {
    cout << "(" << ok << "," << open << "," << close << ")" << endl;
  }
};
string s;
int n;
data T[maxn * 4];
void build(int node, int b, int e) {
  if (b == e) {
    if (s[b] == '(') {
      T[node] = data(0, 1, 0);
    } else {
      T[node] = data(0, 0, 1);
    }
    return;
  }
  build((node << 1), b, ((b + e) >> 1));
  build(((node << 1) | 1), ((b + e) >> 1) + 1, e);
  T[node] = T[(node << 1)] + T[((node << 1) | 1)];
}
data query(int node, int b, int e, int i, int j) {
  if (b > j or e < i) return data(0, 0, 0);
  if (i <= b and e <= j) return T[node];
  return query((node << 1), b, ((b + e) >> 1), i, j) +
         query(((node << 1) | 1), ((b + e) >> 1) + 1, e, i, j);
}
int main() {
  cin >> s;
  n = s.size();
  build(1, 0, n - 1);
  int m;
  cin >> m;
  while (m--) {
    int l, r;
    cin >> l >> r;
    l--, r--;
    cout << query(1, 0, n - 1, l, r).ok * 2 << endl;
  }
}
