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, len;
};
node tree[4 * 1000009];
char str[1000009];
int Q, n;
node merge(node leftChild, node rightChild) {
  node ret;
  int temp = min(leftChild.open, rightChild.close);
  ret.len = leftChild.len + rightChild.len + temp;
  ret.open = leftChild.open + rightChild.open - temp;
  ret.close = leftChild.close + rightChild.close - temp;
  return ret;
}
void build(int idx, int left, int right) {
  if (left == right) {
    if (str[left] == '(')
      tree[idx].open = 1;
    else if (str[left] == ')')
      tree[idx].close = 1;
    tree[idx].len = 0;
    return;
  }
  int mid = (left + right) >> 1;
  build((idx << 1), left, mid);
  build((idx << 1) + 1, mid + 1, right);
  tree[idx] = merge(tree[(idx << 1)], tree[(idx << 1) + 1]);
}
node query(int idx, int left, int right, int i, int j) {
  if (left >= i && right <= j) return tree[idx];
  int mid = (left + right) >> 1;
  if (j <= mid)
    return query((idx << 1), left, mid, i, j);
  else if (i > mid)
    return query((idx << 1) + 1, mid + 1, right, i, j);
  return merge(query((idx << 1), left, mid, i, j),
               query((idx << 1) + 1, mid + 1, right, i, j));
}
int main() {
  scanf("%s", &str);
  n = strlen(str);
  build(1, 0, n - 1);
  scanf("%d", &Q);
  while (Q--) {
    int l, r;
    scanf("%d %d", &l, &r);
    printf("%d\n", query(1, 0, n - 1, l - 1, r - 1).len * 2);
  }
}
