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 N = 1000001;
const int OO = 1e9;
string x;
int n;
struct node {
  int opened = 0;
  int closed = 0;
  int ans = 0;
};
node sg[4 * N];
node mergge(node left, node right) {
  node c;
  int xx = min(left.opened, right.closed);
  c.ans = xx + left.ans + right.ans;
  c.opened = right.opened + left.opened - xx;
  c.closed = left.closed + right.closed - xx;
  return c;
}
node build(int idx, int l, int r) {
  if (l > r) {
    node a;
    return a;
  }
  if (l == r) {
    node a;
    if (x[l] == '(') {
      a.opened++;
    } else
      a.closed++;
    return sg[idx] = a;
  }
  int m = (l + r) / 2;
  node prt1 = build(2 * idx + 1, l, m);
  node prt2 = build(2 * idx + 2, m + 1, r);
  return sg[idx] = mergge(prt1, prt2);
}
node query(int idx, int l, int r, int st, int ed) {
  if (ed < l || st > r) {
    node a;
    return a;
  }
  if (l >= st && r <= ed) return sg[idx];
  int m = (l + r) / 2;
  node prt1 = query(2 * idx + 1, l, m, st, ed);
  node prt2 = query(2 * idx + 2, m + 1, r, st, ed);
  return mergge(prt1, prt2);
}
int main() {
  cin >> x;
  build(0, 0, x.size() - 1);
  int q;
  scanf("%d", &q);
  while (q--) {
    int aa, ab;
    scanf("%d %d", &aa, &ab);
    printf("%d\n", query(0, 0, x.size() - 1, aa - 1, ab - 1).ans * 2);
  }
  return 0;
}
