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;
template <class A, class B>
ostream& operator<<(ostream& o, const pair<A, B>& p) {
  return o << "(" << p.first << ", " << p.second << ")";
}
template <class T>
ostream& operator<<(ostream& o, const vector<T>& v) {
  o << "[";
  for (__typeof((v).end()) it = (v).begin(); it != (v).end(); ++it)
    o << *it << ", ";
  return o << "]";
}
template <class T>
ostream& operator<<(ostream& o, const set<T>& v) {
  o << "[";
  for (__typeof((v).end()) it = (v).begin(); it != (v).end(); ++it)
    o << *it << ", ";
  return o << "]";
}
template <class T>
inline bool read(T& x) {
  int c = getchar();
  int sgn = 1;
  while (~c && c < '0' || c > '9') {
    if (c == '-') sgn = -1;
    c = getchar();
  }
  for (x = 0; ~c && '0' <= c && c <= '9'; c = getchar()) x = x * 10 + c - '0';
  x *= sgn;
  return ~c;
}
const int MX = 4e6 + 7;
int N;
char s[MX];
int tree[MX];
int Ans[MX];
struct Que {
  int st, en, id;
  Que(int _s, int _r, int _i) { st = _s, en = _r, id = _i; }
  Que() {}
};
void update(int idx, int val) {
  while (idx <= N) {
    tree[idx] += val;
    idx += idx & -idx;
  }
}
int query(int idx) {
  int sum = 0;
  while (idx > 0) {
    sum += tree[idx];
    idx -= idx & -idx;
  }
  return sum;
}
int getSum(int l, int r) { return query(r) - query(l - 1); }
bool cmp(Que p, Que q) { return p.en < q.en; }
Que Q[MX];
vector<Que> ret;
int main() {
  scanf("%s", s + 1);
  N = strlen(s + 1);
  stack<int> stk;
  for (__typeof(N) i = (1); i <= N; ++i) {
    if (s[i] == '(') {
      stk.push(i);
      continue;
    }
    if (stk.empty()) continue;
    int st = stk.top();
    stk.pop();
    ret.push_back(Que(st, i, -1));
  }
  sort(ret.begin(), ret.end(), cmp);
  int qq = ({
    int a;
    scanf("%d", &a);
    a;
  });
  for (__typeof(qq) i = (1); i <= qq; ++i) {
    Q[i].st = ({
      int a;
      scanf("%d", &a);
      a;
    });
    Q[i].en = ({
      int a;
      scanf("%d", &a);
      a;
    });
    Q[i].id = i;
  }
  sort(Q + 1, Q + qq + 1, cmp);
  int pos = 0, tot = ret.size();
  for (__typeof(qq) i = (1); i <= qq; ++i) {
    while (pos < tot && ret[pos].en <= Q[i].en) {
      update(ret[pos].st, 1);
      pos++;
    }
    Ans[Q[i].id] = getSum(Q[i].st, Q[i].en);
  }
  for (__typeof(qq) i = (1); i <= qq; ++i) cout << Ans[i] * 2 << endl;
}
