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 = 1e6 + 6;
string s;
struct node_data {
  int open, close, len;
};
struct SegTree {
  node_data t[N << 2];
  node_data combine(const node_data &L, const node_data &R) {
    node_data ret = {L.open + R.open, L.close + R.close, 0};
    ret.len = L.len + R.len + min(L.open - L.len / 2, R.close - R.len / 2) * 2;
    return ret;
  }
  void build(int node, int l, int r) {
    if (l == r) {
      t[node].open = (s[l] == '(');
      t[node].close = (s[l] == ')');
      t[node].len = 0;
      return;
    }
    int mid = l + r >> 1;
    build(node << 1, l, mid);
    build(node << 1 | 1, mid + 1, r);
    t[node] = combine(t[node << 1], t[node << 1 | 1]);
  }
  node_data query(int node, int l, int r, int x, int y) {
    if (x <= l && r <= y) {
      return t[node];
    }
    int mid = l + r >> 1;
    node_data ret = {0, 0, 0};
    if (x <= mid) {
      ret = combine(ret, query(node << 1, l, mid, x, y));
    }
    if (y > mid) {
      ret = combine(ret, query(node << 1 | 1, mid + 1, r, x, y));
    }
    return ret;
  }
};
SegTree tree1;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  int n = s.size();
  int nq;
  cin >> nq;
  tree1.build(1, 0, n - 1);
  while (nq--) {
    int l, r;
    cin >> l >> r;
    --l;
    --r;
    cout << tree1.query(1, 0, n - 1, l, r).len << '\n';
  }
  return 0;
}
