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 long long maxn = 1000100;
long long n, k, a[maxn];
string s;
struct t {
  long long open = 0, close = 0, ans = 0;
} tree[4 * maxn], ft;
t merge(t x, t y) {
  t temp;
  temp.ans = x.ans + y.ans + min(x.open, y.close);
  temp.open = x.open + y.open - min(x.open, y.close);
  temp.close = x.close + y.close - min(x.open, y.close);
  return temp;
}
void build(long long node, long long start, long long end) {
  if (start > end) return;
  if (start == end) {
    if (s[start] == '(')
      tree[node].open++;
    else
      tree[node].close++;
    return;
  }
  long long mid = (start + end) / 2;
  build(2 * node, start, mid);
  build(2 * node + 1, mid + 1, end);
  tree[node] = merge(tree[2 * node], tree[2 * node + 1]);
}
t query(long long node, long long start, long long end, long long l,
        long long r) {
  if (start > end || start > r || l > end) return ft;
  if (l <= start && end <= r) return tree[node];
  long long mid = (start + end) / 2;
  t p1 = query(2 * node, start, mid, l, r);
  t p2 = query(2 * node + 1, mid + 1, end, l, r);
  return merge(p1, p2);
}
void solve() {
  cin >> s;
  n = s.length();
  build(1, 0, n - 1);
  cin >> k;
  while (k--) {
    long long x, y;
    cin >> x >> y;
    x--;
    y--;
    cout << 2 * query(1, 0, n - 1, x, y).ans << '\n';
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  for (long long i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}
