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;
long long M = 1e9 + 7;
long long mpe(long long base, long long exponent, long long modulus) {
  long long result = 1;
  while (exponent > 0) {
    if (exponent % 2 == 1) result = result * base % modulus;
    exponent = exponent >> 1;
    base = base * base % modulus;
  }
  return result;
}
struct nod {
  long long x;
  long long y;
  long long z;
};
const int N = 10 + 1e6;
int n;
int arr[N];
nod tree[N << 1];
string str;
inline nod combine(nod a, nod b) {
  nod res;
  res.x = a.x + b.x + min(a.y, b.z);
  res.y = b.y + a.y - min(a.y, b.z);
  res.z = a.z + b.z - min(a.y, b.z);
  return res;
}
inline void build() {
  for (int i = n; i--;) {
    if (str[i] == '(')
      tree[n + i] = {0, 1, 0};
    else
      tree[n + i] = {0, 0, 1};
  }
  for (int i = n; --i;) tree[i] = combine(tree[i << 1], tree[i << 1 | 1]);
}
inline nod query(int l, int r) {
  nod ansl = {0, 0, 0}, ansr = {0, 0, 0};
  for (l += n, r += n + 1; l < r; l >>= 1, r >>= 1) {
    if (l & 1) ansl = combine(ansl, tree[l++]);
    if (r & 1) ansr = combine(tree[--r], ansr);
  }
  return combine(ansl, ansr);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  ;
  cin >> str;
  n = str.length();
  build();
  long long m;
  cin >> m;
  long long l, r;
  for (long long i = 0; i < m; ++i) {
    cin >> l >> r;
    cout << 2 * query(l - 1, r - 1).x << endl;
  }
  return 0;
}
