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;
string s;
struct {
  int open, close;
} tree[4 * 1000007];
void merge_node(int node) {
  int mn = min(tree[2 * node].open, tree[2 * node + 1].close);
  tree[node].open = tree[2 * node].open + tree[2 * node + 1].open - mn;
  tree[node].close = tree[2 * node].close + tree[2 * node + 1].close - mn;
}
void build(int node, int l, int r) {
  if (l == r) {
    if (s[l - 1] == '(')
      tree[node].open = 1, tree[node].close = 0;
    else
      tree[node].open = 0, tree[node].close = 1;
    return;
  }
  build(2 * node, l, (l + r) / 2);
  build(2 * node + 1, (l + r) / 2 + 1, r);
  merge_node(node);
}
pair<int, int> query(int node, int l, int r, int i, int j) {
  if (l > j or r < i) return pair<int, int>(0, 0);
  if (l >= i and r <= j)
    return pair<int, int>(tree[node].open, tree[node].close);
  pair<int, int> q1 = query(2 * node, l, (l + r) / 2, i, j);
  pair<int, int> q2 = query(2 * node + 1, (l + r) / 2 + 1, r, i, j);
  int mn = min(q1.first, q2.second);
  int retopen = q1.first + q2.first - mn;
  int retclose = q1.second + q2.second - mn;
  return pair<int, int>(retopen, retclose);
}
int main() {
  cin >> s;
  int n = s.size();
  build(1, 1, n);
  int q;
  scanf("%d", &q);
  while (q--) {
    int i, j;
    scanf("%d %d", &i, &j);
    pair<int, int> ans = query(1, 1, n, i, j);
    int out = j - i + 1 - ans.first - ans.second;
    printf("%d\n", out);
  }
}
