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>
#pragma GCC optimize("O3")
using namespace std;
template <typename T, typename K>
inline void umax(T &a, K b) {
  a = max(a, (T)b);
}
template <typename T, typename K>
inline void umin(T &a, K b) {
  a = min(a, (T)b);
}
const int32_t N = 1e6 + 10;
const double EPS = 1e-6;
const int II = 1e9 + 10;
const int MOD = 1e9 + 7;
struct Node {
  int lb, rb, cnt;
} tree[N * 4];
Node sum(const Node &a, const Node &b) {
  Node cur;
  int kek = min(a.lb, b.rb);
  cur.cnt = a.cnt + b.cnt + kek;
  cur.lb = a.lb + b.lb - kek;
  cur.rb = a.rb + b.rb - kek;
  return cur;
}
int n;
string ss;
void build(int v = 1, int tl = 0, int tr = n - 1) {
  if (tl == tr) {
    if (ss[tl] == '(')
      tree[v].lb = 1;
    else
      tree[v].rb = 1;
    return;
  }
  int tm = (tl + tr) >> 1LL;
  build(v << 1, tl, tm);
  build(v << 1 | 1, tm + 1, tr);
  tree[v] = sum(tree[v << 1], tree[v << 1 | 1]);
}
Node getat(int l, int r, int v = 1, int tl = 0, int tr = n - 1) {
  if (tl > r || l > tr) return Node();
  if (l <= tl && tr <= r) return tree[v];
  int tm = (tl + tr) >> 1;
  return sum(getat(l, r, v << 1, tl, tm), getat(l, r, v << 1 | 1, tm + 1, tr));
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> ss;
  n = ss.size();
  build();
  int m;
  cin >> m;
  while (m--) {
    int l, r;
    cin >> l >> r;
    l--;
    r--;
    cout << getat(l, r).cnt * 2 << '\n';
  }
}
