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;
struct node {
  int t, o, c;
};
const int N = 1000001;
int n;
node T[2 * N];
void build() {
  int tmp;
  for (int i = n - 1; i > 0; i--) {
    tmp = min(T[i << 1].o, T[i << 1 | 1].c);
    T[i].t = T[i << 1].t + T[i << 1 | 1].t + (tmp << 1);
    T[i].o = T[i << 1].o + T[i << 1 | 1].o - tmp;
    T[i].c = T[i << 1].c + T[i << 1 | 1].c - tmp;
  }
}
int query(int l, int r) {
  int t = 0, o = 0, c = 0;
  int tmp;
  for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
    if (l & 1) {
      tmp = min(o, T[l].c);
      t += T[l].t + (tmp << 1);
      o += T[l].o - tmp;
      l++;
    }
    if (r & 1) {
      r--;
      tmp = min(c, T[r].o);
      t += T[r].t + (tmp << 1);
      c += T[r].c - tmp;
    }
  }
  t += (min(o, c) << 1);
  return t;
}
int main() {
  int k;
  string brackets;
  cin >> brackets;
  n = brackets.size();
  for (int i = 0; i < n; i++) {
    if (brackets[i] == '(') {
      T[n + i].o = 1;
      T[n + i].c = 0;
    } else {
      T[n + i].c = 1;
      T[n + i].o = 0;
    }
    T[n + i].t = 0;
  }
  build();
  int m;
  cin >> m;
  int x, y;
  for (int i = 0; i < m; i++) {
    cin >> x >> y;
    printf("%d\n", query(x - 1, y));
  }
  return 0;
}
