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 maxn = 1e7 + 7;
int a[maxn], b[maxn], c[maxn], n, q, o, cl, res;
vector<long long> ans;
string s;
void bld(long long ind = 1, long long l = 1, long long r = n) {
  if (l == r) {
    if (s[l - 1] == '(')
      b[ind] = 1;
    else
      c[ind] = 1;
    return;
  }
  long long mid = (l + r) / 2, lc = ind * 2, rc = ind * 2 + 1;
  bld(lc, l, mid);
  bld(rc, mid + 1, r);
  long long t = min(b[lc], c[rc]);
  a[ind] = a[lc] + a[rc] + 2 * t;
  b[ind] = b[rc] + b[lc] - t;
  c[ind] = c[rc] + c[lc] - t;
}
void f(long long x, long long y, long long ind = 1, long long l = 1,
       long long r = n) {
  if (x > r || y < l) return;
  if (y >= r && x <= l) {
    long long t = min(o, c[ind]);
    res += a[ind] + (2 * t);
    o += b[ind] - t;
    cl += c[ind] - t;
    return;
  }
  long long mid = (l + r) / 2;
  f(x, y, ind * 2, l, mid);
  f(x, y, ind * 2 + 1, mid + 1, r);
}
int main() {
  cin >> s;
  n = s.size();
  bld();
  cin >> q;
  for (long long i = 0; i < q; i++) {
    long long l, r;
    cin >> l >> r;
    f(l, r);
    ans.push_back(res);
    res = 0;
    o = 0;
    cl = 0;
  }
  for (auto x : ans) cout << x << endl;
  return 0;
}
