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;
bool isPowerOfTwo(long long n) {
  if (n == 0) return false;
  return (ceil(log2(n)) == floor(log2(n)));
}
long long power(long long x, unsigned long long y) {
  long long res = 1;
  x = x;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
bool isprime(long long x) {
  if (x < 2) return false;
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) return false;
  }
  return true;
}
vector<long long> primeFactors(long long n) {
  vector<long long> v;
  while (n % 2 == 0) {
    v.push_back(2);
    n = n / 2;
  }
  for (long long i = 3; i <= sqrt(n); i = i + 2) {
    while (n % i == 0) {
      v.push_back(i);
      n = n / i;
    }
  }
  if (n > 2) v.push_back(n);
  return v;
}
long long ans[4000005], o[1000001], c[1000001];
void segtree(long long x, long long l, long long r) {
  if (l == r) {
    ans[x] = o[l];
    return;
  }
  long long m = (l + r) / 2;
  segtree(2 * x + 1, l, m);
  segtree(2 * x + 2, m + 1, r);
  ans[x] = min(ans[2 * x + 1], ans[2 * x + 2]);
}
long long query(long long x, long long l, long long r, long long l1,
                long long r1) {
  if (l >= l1 && r <= r1) return ans[x];
  if (r < l1 || l > r1) return INT_MAX;
  long long m = (l + r) / 2;
  return min(query(2 * x + 1, l, m, l1, r1),
             query(2 * x + 2, m + 1, r, l1, r1));
}
void solve() {
  string s;
  long long q;
  cin >> s >> q;
  long long n = s.size();
  o[0] = 0;
  c[0] = 0;
  for (long long i = 0; i < n; i++) {
    o[i + 1] = o[i];
    if (s[i] == '(')
      o[i + 1]++;
    else
      o[i + 1]--;
    c[i + 1] = c[i];
    if (s[i] == ')') c[i + 1]++;
  }
  segtree(0, 1, n);
  while (q--) {
    long long x, y;
    cin >> x >> y;
    long long k;
    k = 2 *
        ((c[y] - c[x - 1]) + (min(query(0ll, 1ll, n, x, y) - o[x - 1], 0ll)));
    cout << k << endl;
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  t = 1;
  while (t--) {
    solve();
  }
}
