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 BIT {
  int n;
  vector<int> a;
  BIT(int _n) {
    n = _n;
    a = vector<int>(n + 1, 0);
  }
  int sum(int k) {
    int ans = 0;
    for (int idx = k; idx > 0; idx -= (idx & -idx)) ans += a[idx];
    return ans;
  }
  int query(int l, int r) { return sum(r) - sum(l - 1); }
  void update(int k, int v) {
    for (int idx = k; idx <= n; idx += (idx & -idx)) a[idx] += v;
  }
};
struct intv {
  int l, r, id;
  intv() {}
  intv(int _l, int _r) : l(_l), r(_r) {}
  bool operator<(const intv& rhs) const { return r < rhs.r; }
};
const int N = 1000000 + 10;
char s[N];
stack<int> lb;
vector<intv> p, q;
vector<int> ans;
int main() {
  scanf("%s", s + 1);
  int n = strlen(s + 1);
  for (int i = 1; i <= n; i++) {
    if (s[i] == '(') {
      lb.push(i);
    } else if (lb.size()) {
      p.push_back(intv(lb.top(), i));
      lb.pop();
    }
  }
  int m;
  scanf("%d", &m);
  q = vector<intv>(m);
  ans = vector<int>(m);
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &q[i].l, &q[i].r);
    q[i].id = i;
  }
  sort(q.begin(), q.end());
  BIT bit(n);
  int qTop = 0, pTop = 0;
  for (int i = 1; i <= n; i++) {
    for (; pTop < p.size() && p[pTop].r == i; pTop++) {
      const intv& top = p[pTop];
      bit.update(top.l, 2);
    }
    for (; qTop < m && q[qTop].r == i; qTop++) {
      const intv& top = q[qTop];
      ans[top.id] = bit.query(top.l, top.r);
    }
  }
  for (int i = 0; i < m; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}
