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;
int const MAXN = 4000000;
struct node {
  int o, c, t;
};
node tree[MAXN];
string s;
void build(int i, int l, int r) {
  if (l == r) {
    tree[i].o = (s[l] == '('), tree[i].c = (s[l] == ')'), tree[i].t = 0;
  } else {
    int mid = (l + r) / 2;
    build(2 * i, l, mid);
    build(2 * i + 1, mid + 1, r);
    int temp = min(tree[2 * i].o, tree[2 * i + 1].c);
    tree[i].t = tree[2 * i].t + tree[2 * i + 1].t + temp * 2;
    tree[i].o = tree[2 * i].o + tree[2 * i + 1].o - temp;
    tree[i].c = tree[2 * i].c + tree[2 * i + 1].c - temp;
  }
}
node query(int index, int l, int r, int i, int j) {
  if (i <= l && j >= r) {
    return tree[index];
  }
  int mid = (l + r) / 2;
  if (j <= mid) return query(index * 2, l, mid, i, j);
  if (i > mid) return query(index * 2 + 1, mid + 1, r, i, j);
  node a = query(index * 2, l, mid, i, mid);
  node b = query(index * 2 + 1, mid + 1, r, mid + 1, j);
  int temp = min(a.o, b.c);
  node p;
  p.o = a.o + b.o - temp;
  p.c = a.c + b.c - temp;
  p.t = a.t + b.t + temp * 2;
  return p;
}
int main() {
  (ios::sync_with_stdio(false));
  int m;
  cin >> s;
  int n = s.length();
  build(1, 0, n - 1);
  cin >> m;
  for (int i = 0; i < (m); ++i) {
    int l, r;
    cin >> l >> r;
    node ans = query(1, 0, n - 1, --l, --r);
    cout << ans.t << '\n';
  }
  return 0;
}
