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;
char a[1000001];
int n;
struct node {
  int ans, open, close;
} t[4000001], we;
void build(int v = 1, int tl = 1, int tr = n) {
  if (tl == tr) {
    if (a[tl] == '(') {
      t[v].open++;
    } else
      t[v].close++;
    return;
  }
  int tm = (tl + tr) / 2;
  build(v + v, tl, tm);
  build(v + v + 1, tm + 1, tr);
  int qwe = min(t[v + v].open, t[v + v + 1].close);
  t[v].ans = t[v + v].ans + t[v + v + 1].ans + qwe;
  t[v].open = t[v + v].open + t[v + v + 1].open - qwe;
  t[v].close = t[v + v].close + t[v + v + 1].close - qwe;
}
node get(int l, int r, int v = 1, int tl = 1, int tr = n) {
  if (tr < l || tl > r) return we;
  if (tl >= l && tr <= r) return t[v];
  int tm = (tl + tr) / 2;
  node x = get(l, r, v + v, tl, tm);
  node y = get(l, r, v + v + 1, tm + 1, tr);
  int qwe = min(x.open, y.close);
  x.ans = x.ans + y.ans + qwe;
  x.open = x.open + y.open - qwe;
  x.close = x.close + y.close - qwe;
  return x;
}
int main() {
  string s;
  cin >> s;
  n = s.size();
  for (int i = 1; i <= n; i++) {
    a[i] = s[i - 1];
  }
  build();
  int q;
  cin >> q;
  while (q--) {
    int x, y;
    cin >> x >> y;
    cout << get(x, y).ans * 2 << endl;
    ;
  }
  return 0;
}
