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;
string s;
string t;
int m;
typedef struct node {
  int open;
  int close;
  int rez;
} NODE;
NODE tree[4000001];
void build(int pos, int l, int r) {
  if (l == r) {
    tree[pos] = {0, 0, 0};
    if (s[l] == '(')
      tree[pos].open++;
    else
      tree[pos].close++;
    return;
  }
  int mid = (l + r) / 2;
  build(2 * pos, l, mid);
  build(2 * pos + 1, mid + 1, r);
  tree[pos] = {0, 0, 0};
  NODE a = tree[2 * pos];
  NODE b = tree[2 * pos + 1];
  int buff = min(a.open, b.close);
  tree[pos].open = a.open + b.open - buff;
  tree[pos].close = a.close + b.close - buff;
  tree[pos].rez = a.rez + b.rez + 2 * buff;
}
NODE query(int pos, int l, int r, int ll, int rr) {
  if (ll > r || rr < l) return {0, 0, 0};
  if (l >= ll && r <= rr) return tree[pos];
  int mid = (l + r) / 2;
  NODE a = query(2 * pos, l, mid, ll, rr);
  NODE b = query(2 * pos + 1, mid + 1, r, ll, rr);
  NODE c = {0, 0, 0};
  c.open = a.open + b.open - min(a.open, b.close);
  c.close = a.close + b.close - min(a.open, b.close);
  c.rez = a.rez + b.rez + 2 * min(a.open, b.close);
  return c;
}
int main() {
  cin >> s;
  build(1, 0, s.size() - 1);
  cin >> m;
  for (int i = 0; i < m; i++) {
    int x;
    cin >> x;
    int y;
    cin >> y;
    cout << query(1, 0, s.size() - 1, x - 1, y - 1).rez << endl;
  }
  return 0;
}
