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 len, open, close;
  Node() {
    len = 0;
    open = 0;
    close = 0;
  }
};
int N, Q;
string stuff;
Node tree[4000001];
int left(int p) { return p << 1; }
int right(int p) { return (p << 1) + 1; }
void build(int p, int L, int R) {
  if (L == R) {
    if (stuff[L] == '(')
      tree[p].open = 1;
    else
      tree[p].close = 1;
    return;
  }
  int li = left(p);
  int ri = right(p);
  build(li, L, (L + R) / 2);
  build(ri, (L + R) / 2 + 1, R);
  int t = min(tree[li].open, tree[ri].close);
  tree[p].len = tree[li].len + tree[ri].len + 2 * t;
  tree[p].open = tree[li].open + tree[ri].open - t;
  tree[p].close = tree[li].close + tree[ri].close - t;
}
Node query(int p, int L, int R, int i, int j) {
  Node ans, a1, a2;
  if (j < L || i > R) return ans;
  if (i <= L && j >= R) return tree[p];
  int li = left(p);
  int ri = right(p);
  a1 = query(li, L, (L + R) / 2, i, j);
  a2 = query(ri, (L + R) / 2 + 1, R, i, j);
  int t = min(a1.open, a2.close);
  ans.len = a1.len + a2.len + 2 * t;
  ans.open = a1.open + a2.open - t;
  ans.close = a1.close + a2.close - t;
  return ans;
}
int main() {
  cin >> stuff;
  N = stuff.length();
  scanf("%d", &Q);
  build(1, 0, N - 1);
  for (int i = 0; i < Q; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    cout << query(1, 0, N - 1, a - 1, b - 1).len << endl;
  }
  return 0;
}
