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;
int dx[] = {0, 1, -1, 0};
int dy[] = {1, 0, 0, -1};
long double PI = 3.14159265358979323846264338327950;
const int N = (1e6 + 10) * 4, M = 1e7;
struct Node {
  int open, close, mx;
};
Node SegmentTree[N];
string s;
int n;
void build(int st = 0, int end = n - 1, int root = 1) {
  if (st == end) {
    SegmentTree[root].open = (s[st] == '(');
    SegmentTree[root].close = (s[st] == ')');
    SegmentTree[root].mx = 0;
    return;
  }
  build(st, ((end + st) >> 1), (root << 1));
  build(((end + st) >> 1) + 1, end, ((root << 1) | 1));
  Node x = SegmentTree[(root << 1)], y = SegmentTree[((root << 1) | 1)];
  Node& ret = SegmentTree[root];
  int add = min(x.open, y.close);
  ret.mx = x.mx + y.mx + add;
  ret.open = x.open - add + y.open;
  ret.close = y.close - add + x.close;
}
Node query(int l, int r, int st = 0, int end = n - 1, int root = 1) {
  Node ret;
  ret.open = ret.mx = ret.close = 0;
  if (l > end || r < st) {
    return ret;
  }
  if (st >= l && end <= r) {
    return SegmentTree[root];
  }
  Node x = query(l, r, st, ((end + st) >> 1), (root << 1)),
       y = query(l, r, ((end + st) >> 1) + 1, end, ((root << 1) | 1));
  int add = min(x.open, y.close);
  ret.mx = x.mx + y.mx + add;
  ret.open = x.open - add + y.open;
  ret.close = y.close - add + x.close;
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> s;
  n = (int)s.size();
  build();
  int q;
  cin >> q;
  while (q--) {
    int l, r;
    cin >> l >> r;
    l--, r--;
    cout << (query(l, r).mx << 1) << "\n";
  }
}
