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 nt {
  int a, b, c;
};
char a[5000000];
string s;
nt tree[5000000];
int n, sz, m;
nt c;
nt comb(nt a, nt b) {
  int t = min(a.b, b.c);
  c.a = a.a + b.a + t;
  c.b = a.b + b.b - t;
  c.c = a.c + b.c - t;
  return c;
}
void build(int v, int l, int r) {
  if (l == r) {
    if (a[l] == '(')
      tree[v].b = 1;
    else if (a[l] == ')')
      tree[v].c = 1;
    return;
  }
  int m = (l + r) / 2;
  build(v + v, l, m);
  build(v + v + 1, m + 1, r);
  tree[v] = comb(tree[v + v], tree[v + v + 1]);
}
nt get(int v, int l, int r, int ll, int rr) {
  if (l == ll && r == rr) return tree[v];
  int m = (l + r) / 2;
  if (m >= rr)
    return get(v + v, l, m, ll, rr);
  else if (m < ll)
    return get(v + v + 1, m + 1, r, ll, rr);
  else
    return comb(get(v + v, l, m, ll, m), get(v + v + 1, m + 1, r, m + 1, rr));
}
int main() {
  ios::sync_with_stdio(false);
  cin >> s;
  n = s.size();
  for (int i = 0; i < n; ++i) a[i] = s[i];
  sz = 1;
  while (sz < n) sz += sz;
  build(1, 0, sz - 1);
  cin >> m;
  for (int i = 0; i < m; ++i) {
    int l, r;
    cin >> l >> r;
    l--;
    r--;
    cout << get(1, 0, sz - 1, l, r).a * 2 << endl;
  }
  return 0;
}
