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 N = 1e6 + 7;
struct node {
  int l = -1, r = -1;
  int a = 0, b = 0, c = 0;
};
string s;
vector<node> t(4 * N);
void build(int l, int r, int v = 1) {
  t[v].l = l;
  t[v].r = r;
  if (l == r) {
    if (s[l] == '(') {
      t[v].a = 0, t[v].b = 1, t[v].c = 0;
    } else {
      t[v].a = 0, t[v].b = 0, t[v].c = 1;
    }
    return;
  }
  int m = (l + r) / 2;
  build(l, m, 2 * v);
  build(m + 1, r, 2 * v + 1);
  int d = min(t[2 * v].b, t[2 * v + 1].c);
  t[v].a = t[2 * v].a + t[2 * v + 1].a + 2 * d;
  t[v].b = t[2 * v].b + t[2 * v + 1].b - d;
  t[v].c = t[2 * v].c + t[2 * v + 1].c - d;
}
node neitral;
node get(int l, int r, int v = 1) {
  if (t[v].l > r || t[v].r < l) return neitral;
  if (l <= t[v].l && t[v].r <= r) return t[v];
  node left = get(l, r, 2 * v);
  node right = get(l, r, 2 * v + 1);
  node ans;
  int d = min(left.b, right.c);
  ans.a = left.a + right.a + 2 * d;
  ans.b = left.b + right.b - d;
  ans.c = left.c + right.c - d;
  return ans;
}
int main() {
  ios ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  int m;
  cin >> m;
  build(0, (int)s.size() - 1);
  for (int i = 0; i < m; i++) {
    int l, r;
    cin >> l >> r;
    l--, r--;
    cout << get(l, r).a << '\n';
  }
}
