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 inf = 1e18;
long long fast_pow(long long base, long long n, long long M = inf) {
  if (n == 0) return 1;
  if (n == 1) return base;
  long long halfn = fast_pow(base, n / 2, M);
  if (n % 2 == 0)
    return (halfn * halfn) % M;
  else
    return (((halfn * halfn) % M) * base) % M;
}
struct S {
  long long t, ul, ur;
  S(long long x = 0, long long y = 0, long long z = 0) {
    t = x, ul = y, ur = z;
  }
};
S a[4000005];
S add(S a, S b) {
  S c = {0, 0, 0};
  long long l = min(a.ul, b.ur);
  c.t = a.t + b.t + 2 * l;
  c.ul = a.ul + b.ul - l;
  c.ur = a.ur + b.ur - l;
  return c;
}
void build(string &s, long long node, long long l, long long r) {
  if (l == r) {
    if (s[l] == '(')
      a[node].ul++;
    else
      a[node].ur++;
  } else {
    long long m = (l + r) / 2;
    build(s, 2 * node, l, m);
    build(s, 2 * node + 1, 1 + m, r);
    a[node] = add(a[2 * node], a[2 * node + 1]);
  }
}
S query(long long node, long long l, long long r, long long x, long long y) {
  if (x > r || y < l) return {0, 0, 0};
  if (x <= l && r <= y) return a[node];
  long long m = (l + r) / 2;
  return add(query(2 * node, l, m, x, y), query(2 * node + 1, m + 1, r, x, y));
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  string s;
  cin >> s;
  long long i, q, n = s.size();
  build(s, 1, 0, n - 1);
  cin >> q;
  while (q--) {
    long long x, y;
    cin >> x >> y;
    cout << query(1, 0, n - 1, x - 1, y - 1).t << "\n";
  }
}
