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 = 1000 * 1000 + 20;
struct node {
  int l, r, cntO, cntC, ans;
  node() : l(-1), r(-1), ans(0), cntO(0), cntC(0) {}
} seg[N << 2];
int n, q;
string s;
node Merge(node lc, node rc) {
  node res;
  int mn = min(lc.cntO, rc.cntC);
  res.ans = lc.ans + rc.ans + 2 * mn;
  res.cntO = lc.cntO + rc.cntO - mn;
  res.cntC = lc.cntC + rc.cntC - mn;
  return res;
}
void build(int id = 1, int l = 0, int r = n) {
  if (l == r) {
    seg[id].cntO = s[r] == '(' ? 1 : 0;
    seg[id].cntC = s[r] == ')' ? 1 : 0;
    seg[id].l = l, seg[id].r = r;
    return;
  }
  int mid = l + r >> 1;
  build(id << 1, l, mid);
  build(id << 1 | 1, mid + 1, r);
  seg[id] = Merge(seg[id << 1], seg[id << 1 | 1]);
  seg[id].l = l, seg[id].r = r;
}
node get(int lq, int rq, int id = 1) {
  int l = seg[id].l, r = seg[id].r;
  if (lq > r || l > rq) return node();
  if (l >= lq && r <= rq) return seg[id];
  int mid = l + r >> 1;
  return Merge(get(lq, rq, id << 1), get(lq, rq, id << 1 | 1));
}
int main() {
  ios::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  cin >> s >> q;
  n = s.size();
  build();
  while (q--) {
    int l, r;
    cin >> l >> r;
    cout << get(--l, --r).ans << '\n';
  }
  return 0;
}
