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 MAX = 1e6 + 10;
string str;
struct node {
  int mx, lt, rt;
  node() { mx = lt = rt = 0; }
};
node seg[MAX << 2];
void build(int nd, int l, int r) {
  if (l == r) {
    seg[nd] = node();
    if (str[l] == '(')
      seg[nd].rt = 1;
    else
      seg[nd].lt = 1;
    return;
  }
  int mid = (l + r) / 2;
  build(nd * 2, l, mid);
  build(nd * 2 + 1, mid + 1, r);
  int tmp = min(seg[nd * 2].rt, seg[nd * 2 + 1].lt);
  seg[nd].mx = seg[nd * 2].mx + seg[nd * 2 + 1].mx + tmp;
  seg[nd].rt = seg[nd * 2].rt + seg[nd * 2 + 1].rt - tmp;
  seg[nd].lt = seg[nd * 2].lt + seg[nd * 2 + 1].lt - tmp;
}
node query(int nd, int l, int r, int a, int b) {
  if (a <= l && r <= b) return seg[nd];
  node ret = node();
  if (b < l || r < a) return ret;
  int mid = (l + r) / 2;
  node stPath = query(nd * 2, l, mid, a, b);
  node ndPath = query(nd * 2 + 1, mid + 1, r, a, b);
  int tmp = min(stPath.rt, ndPath.lt);
  ret.mx = stPath.mx + ndPath.mx + tmp;
  ret.rt = stPath.rt + ndPath.rt - tmp;
  ret.lt = stPath.lt + ndPath.lt - tmp;
  return ret;
}
void solve() {
  cin >> str;
  str = " " + str;
  int n = str.size();
  build(1, 1, n);
  int q;
  cin >> q;
  while (q--) {
    int a, b;
    cin >> a >> b;
    cout << 2 * query(1, 1, n, a, b).mx << endl;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  if (false) cin >> t;
  while (t--) {
    solve();
  }
}
