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 = 1000010;
string s;
class node {
 public:
  int pairs, open, close;
} tree[4 * N];
node merge(node l, node r) {
  node ret;
  ret.pairs = l.pairs + r.pairs + min(l.open, r.close);
  ret.open = l.open + r.open - min(l.open, r.close);
  ret.close = l.close + r.close - min(l.open, r.close);
  return ret;
}
void build(int root, int start, int end) {
  if (start == end) {
    tree[root].pairs = 0;
    tree[root].open = 1, tree[root].close = 0;
    if (s[start] == ')') swap(tree[root].open, tree[root].close);
    return;
  }
  int mid = (start + end) >> 1;
  build(2 * root, start, mid);
  build(2 * root + 1, mid + 1, end);
  tree[root] = merge(tree[2 * root], tree[2 * root + 1]);
}
node query(int root, int i, int j, int start, int end) {
  if (i <= start && j >= end) return tree[root];
  int mid = (start + end) >> 1;
  if (j <= mid)
    return query(2 * root, i, j, start, mid);
  else if (i > mid)
    return query(2 * root + 1, i, j, mid + 1, end);
  else
    return merge(query(2 * root, i, j, start, mid),
                 query(2 * root + 1, i, j, mid + 1, end));
}
int main() {
  cin >> s;
  int len = s.length();
  build(1, 0, len - 1);
  int m, x, y;
  cin >> m;
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &x, &y);
    printf("%d\n", query(1, x - 1, y - 1, 0, len - 1).pairs * 2);
  }
}
