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 seg {
  int left;
  int right;
  int total;
};
seg segtree[3000015];
string s;
int n;
void make_tree(int node, int l, int r) {
  if (l == r) {
    if (s[l - 1] == '(') {
      segtree[node].left = 1;
      segtree[node].right = segtree[node].total = 0;
    } else {
      segtree[node].right = 1;
      segtree[node].left = segtree[node].left = 0;
    }
    return;
  }
  int mid = (l + r) >> 1;
  int L = node << 1;
  int R = (node << 1) + 1;
  make_tree(L, l, mid);
  make_tree(R, mid + 1, r);
  int n = min(segtree[L].left, segtree[R].right);
  segtree[node].left = segtree[L].left + segtree[R].left - n;
  segtree[node].right = segtree[R].right + segtree[L].right - n;
  segtree[node].total = segtree[L].total + segtree[R].total + n;
}
seg query(int node, int l, int r, int L, int R) {
  if (l > R || r < L) {
    seg a;
    a.left = a.right = a.total = 0;
    return a;
  }
  if (l >= L && R >= r) {
    return segtree[node];
  }
  int mid = (l + r) >> 1;
  int left = node << 1;
  int right = left + 1;
  seg p = query(left, l, mid, L, R);
  seg q = query(right, mid + 1, r, L, R);
  int n = min(p.left, q.right);
  p.total = p.total + q.total + n;
  p.left = p.left + q.left - n;
  p.right = p.right + q.right - n;
  return p;
}
int main() {
  cin >> s;
  n = s.size();
  make_tree(1, 1, n);
  int q;
  cin >> q;
  while (q--) {
    int a, b;
    cin >> a >> b;
    seg q = query(1, 1, n, a, b);
    cout << 2 * q.total << endl;
  }
}
