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;
int n, m, per[1000100], sol[100100], aib[1000100];
char s[1000100];
int St[1000100], vf;
struct Intrebare {
  int st, dr, ind;
  bool operator<(const Intrebare &A) const { return dr < A.dr; }
};
Intrebare Q[100100];
inline void Aduna(int poz) {
  while (poz <= n) {
    aib[poz]++;
    poz += (poz & -poz);
  }
}
inline int Suma(int poz) {
  int sum = 0;
  while (poz > 0) {
    sum += aib[poz];
    poz -= (poz & -poz);
  }
  return sum;
}
int main() {
  int i, j;
  cin >> (s + 1);
  n = strlen(s + 1);
  for (i = 1; i <= n; ++i) {
    if (s[i] == '(')
      St[++vf] = i;
    else {
      if (vf > 0) {
        per[i] = St[vf];
        vf--;
      }
    }
  }
  cin >> m;
  for (i = 1; i <= m; ++i) {
    cin >> Q[i].st >> Q[i].dr;
    Q[i].ind = i;
  }
  sort(Q + 1, Q + m + 1);
  for (i = j = 1; i <= n && j <= m; ++i) {
    if (per[i]) Aduna(per[i]);
    while (j <= m && Q[j].dr == i) {
      sol[Q[j].ind] = 2 * (Suma(i) - Suma(Q[j].st - 1));
      j++;
    }
  }
  for (i = 1; i <= m; ++i) cout << sol[i] << "\n";
  return 0;
}
