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>
long long max(long long x, long long y) { return x > y ? x : y; }
long long min(long long x, long long y) { return x < y ? x : y; }
using namespace std;
const long long INF = 1e18 + 14;
const long long N = 4e6 + 9;
string s;
long long n, m;
long long op[N], cl[N];
struct node {
  long long g = 0, c = 0, o = 0;
} t[N];
void build(long long at, long long l, long long r) {
  if (l == r) {
    if (s[l] == '(')
      t[at].o = 1;
    else
      t[at].c = 1;
    return;
  }
  long long mid = (l + r) / 2;
  build(at * 2 + 1, l, mid);
  build(at * 2 + 2, mid + 1, r);
  t[at].g =
      t[at * 2 + 1].g + t[at * 2 + 2].g + min(t[at * 2 + 1].o, t[at * 2 + 2].c);
  t[at].o = max(0, t[at * 2 + 1].o - t[at * 2 + 2].c) + t[at * 2 + 2].o;
  t[at].c = max(0, t[at * 2 + 2].c - t[at * 2 + 1].o) + t[at * 2 + 1].c;
}
long long query(long long at, long long l, long long r, long long L,
                long long R) {
  if (l == L && r == R) return t[at].g;
  long long mid = (L + R) / 2, lres = 0, rres = 0;
  if (l > mid) return query(at * 2 + 2, l, r, mid + 1, R);
  if (r <= mid) return query(at * 2 + 1, l, r, L, mid);
  lres = query(at * 2 + 1, l, mid, L, mid);
  rres = query(at * 2 + 2, mid + 1, r, mid + 1, R);
  long long ores = (op[mid] - op[l - 1]) - lres;
  long long cres = (cl[r] - cl[mid]) - rres;
  return lres + rres + min(ores, cres);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> s;
  n = s.size();
  build(0, 0, n - 1);
  for (long long i = 0; i < n; i++) {
    op[i] = op[i - 1] + (s[i] == '(');
    cl[i] = cl[i - 1] + (s[i] == ')');
  }
  cin >> m;
  for (long long i = 0; i < m; i++) {
    long long l, r;
    cin >> l >> r;
    l--;
    r--;
    cout << query(0, l, r, 0, n - 1) * 2 << endl;
  }
  return 0;
}
