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 open, close, result;
};
Node tree[3000040];
char s[1000010];
int length, q, strt, nd;
void Build(int l, int r, int index) {
  if (l == r) {
    tree[index].open = (s[r - 1] == '(');
    tree[index].close = (s[r - 1] == ')');
    tree[index].result = 0;
    return;
  }
  Build(l, (l + r) / 2, index * 2);
  Build((l + r) / 2 + 1, r, index * 2 + 1);
  int x = min(tree[index * 2].open, tree[index * 2 + 1].close);
  tree[index].result =
      x * 2 + tree[index * 2].result + tree[index * 2 + 1].result;
  tree[index].open = tree[index * 2].open + tree[index * 2 + 1].open - x;
  tree[index].close = tree[index * 2].close + tree[index * 2 + 1].close - x;
}
Node solve(int l, int r, int i) {
  Node node;
  if (l > nd || r < strt) {
    node.open = 0, node.close = 0, node.result = 0;
    return node;
  }
  if (l >= strt && r <= nd) return tree[i];
  Node a = solve(l, (l + r) / 2, i * 2),
       b = solve((l + r) / 2 + 1, r, i * 2 + 1);
  int x = min(a.open, b.close);
  node.result = a.result + b.result + x * 2;
  node.close = a.close + b.close - x;
  node.open = a.open + b.open - x;
  return node;
}
int main() {
  scanf("%s", s);
  scanf("%d", &q);
  length = strlen(s);
  Build(1, length, 1);
  for (int i = 0; i < q; i++) {
    scanf("%d%d", &strt, &nd);
    printf("%d\n", solve(1, length, 1).result);
  }
  return 0;
}
