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;
struct Node {
  int close, open, score;
};
class Task {
 private:
  string s;
  int m;
  vector<pair<int, int>> queries;
  int n;
  vector<Node> sg_tree;
  void Read() {
    cin >> s >> m;
    queries.resize(m);
    for (auto &[l, r] : queries) {
      cin >> l >> r;
    }
    n = int(s.size());
    s = " " + s;
  }
  void Solve() {
    sg_tree.resize(4 * n + 4);
    Build(1, 1, n);
    for (auto &[l, r] : queries) {
      cout << 2 * Get(1, 1, n, l, r).score << "\n";
    }
  }
  void Build(int tv, int tl, int tr) {
    if (tl == tr) {
      sg_tree[tv] = Node{s[tl] == ')', s[tl] == '(', 0};
    } else {
      int tm = ((tl + tr) >> 1), lc = 2 * tv, rc = lc + 1;
      Build(lc, tl, tm);
      Build(rc, tm + 1, tr);
      sg_tree[tv] = Merge(sg_tree[lc], sg_tree[rc]);
    }
    42;
  }
  Node Get(int tv, int tl, int tr, int ql, int qr) {
    if (qr < tl || ql > tr) {
      return Node{0, 0, 0};
    } else if (ql <= tl && tr <= qr) {
      return sg_tree[tv];
    } else {
      int tm = ((tl + tr) >> 1), lc = 2 * tv, rc = lc + 1;
      return Merge(Get(lc, tl, tm, ql, qr), Get(rc, tm + 1, tr, ql, qr));
    }
  }
  Node Merge(Node a, Node b) {
    return Node{a.close + max(0, b.close - a.open),
                b.open + max(0, a.open - b.close),
                a.score + b.score + min(a.open, b.close)};
  }

 public:
  void Perform() {
    Read();
    Solve();
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  Task t;
  t.Perform();
  return 0;
}
