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;
const int maxn = 1e6 + 5;
int n, m;
char a[maxn];
struct data {
  int open, close, all;
  data() {
    open = 0;
    close = 0;
    all = 0;
  }
  data(int _open, int _close, int _all) {
    open = _open;
    close = _close;
    all = _all;
  }
  friend data merge(const data &a, const data &b) {
    data ret;
    int match = min(a.open, b.close);
    ret.open = a.open + b.open - match;
    ret.close = a.close + b.close - match;
    ret.all = a.all + b.all + match;
    return ret;
  }
} seg[maxn << 2];
void init(int k, int b, int e) {
  if (b == e) {
    if (a[b] == '(')
      seg[k].open = 1;
    else
      seg[k].close = 1;
    return;
  }
  int m = (b + e) >> 1;
  init(k + k, b, m);
  init(k + k + 1, m + 1, e);
  seg[k] = merge(seg[k + k], seg[k + k + 1]);
}
data find(int k, int b, int e, int x, int y) {
  if (b > y or e < x) return data();
  if (b >= x and e <= y) return seg[k];
  int m = (b + e) >> 1;
  data l = find(k + k, b, m, x, y);
  data r = find(k + k + 1, m + 1, e, x, y);
  return merge(l, r);
}
int main() {
  scanf("%s", a + 1);
  n = strlen(a + 1);
  scanf("%d", &m);
  init(1, 1, n);
  for (int i = 1, l, r; i <= m; i++) {
    scanf("%d%d", &l, &r);
    printf("%d\n", find(1, 1, n, l, r).all << 1);
  }
  return 0;
}
