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 INF = 1e9 + 1;
struct Point {
  int len, o, c;
  Point() : len(-1), o(-1), c(-1) {}
  Point(int len, int o, int c) : len(len), o(o), c(c) {}
};
vector<Point> t;
string s;
void build(int ver, int l, int r) {
  if (r - l == 1) {
    if (s[l] == '(')
      t[ver] = Point(0, 1, 0);
    else if (s[l] == ')')
      t[ver] = Point(0, 0, 1);
  } else {
    int m = l + (r - l) / 2;
    build(ver * 2, l, m);
    build(ver * 2 + 1, m, r);
    Point a = t[ver * 2], b = t[ver * 2 + 1];
    if (a.len == -1 and b.len == -1)
      t[ver] = Point(-1, -1, -1);
    else if (a.len == -1)
      t[ver] = b;
    else if (b.len == -1)
      t[ver] = a;
    else {
      int cnt = min(t[ver * 2].o, t[ver * 2 + 1].c);
      t[ver] = Point(a.len + b.len + 2 * cnt, a.o + b.o - cnt, a.c + b.c - cnt);
    }
  }
}
Point get(int ver, int l, int r, int L, int R) {
  if (r <= L or R <= l)
    return Point(-1, -1, -1);
  else if (l == L and r == R)
    return t[ver];
  else {
    int m = l + (r - l) / 2;
    Point a = get(ver * 2, l, m, L, min(R, m));
    Point b = get(ver * 2 + 1, m, r, max(m, L), R);
    if (a.len == -1 and b.len == -1)
      return Point(-1, -1, -1);
    else if (a.len == -1)
      return b;
    else if (b.len == -1)
      return a;
    else {
      int cnt = min(a.o, b.c);
      return Point(a.len + b.len + 2 * cnt, a.o + b.o - cnt, a.c + b.c - cnt);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  t.resize(s.size() * 4);
  build(1, 0, s.size());
  int n;
  cin >> n;
  int l, r;
  for (int i = 0; i < n; ++i) {
    cin >> l >> r;
    cout << get(1, 0, s.size(), l - 1, r).len << "\n";
  }
}
