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;
char s[1000010];
struct Tree {
  int sl, sr, sum1, sum2, tot;
} tree[2 * 1000010];
struct Que {
  int sum1, sum2, tot;
};
int n, o, top, a, b;
void build(int &t, int l, int r) {
  t = ++top;
  if (l == r) {
    if (s[l] == '(')
      tree[t].sum1 = 1;
    else
      tree[t].sum2 = 1;
    return;
  }
  int mid = (l + r) / 2;
  build(tree[t].sl, l, mid);
  build(tree[t].sr, mid + 1, r);
  tree[t].tot = tree[tree[t].sl].tot + tree[tree[t].sr].tot +
                min(tree[tree[t].sl].sum1, tree[tree[t].sr].sum2);
  tree[t].sum1 = tree[tree[t].sl].sum1 + tree[tree[t].sr].sum1 -
                 min(tree[tree[t].sl].sum1, tree[tree[t].sr].sum2);
  tree[t].sum2 = tree[tree[t].sl].sum2 + tree[tree[t].sr].sum2 -
                 min(tree[tree[t].sl].sum1, tree[tree[t].sr].sum2);
}
Que find(int t, int l, int r) {
  if (a <= l && b >= r) return (Que){tree[t].sum1, tree[t].sum2, tree[t].tot};
  int mid = (l + r) / 2;
  if (b <= mid) return find(tree[t].sl, l, mid);
  if (a > mid) return find(tree[t].sr, mid + 1, r);
  Que t1, t2, ans;
  t1 = find(tree[t].sl, l, mid);
  t2 = find(tree[t].sr, mid + 1, r);
  ans.tot = t1.tot + t2.tot + min(t1.sum1, t2.sum2);
  ans.sum1 = t1.sum1 + t2.sum1 - min(t1.sum1, t2.sum2);
  ans.sum2 = t1.sum2 + t2.sum2 - min(t1.sum1, t2.sum2);
  return ans;
}
int main() {
  scanf("%s", &s[1]);
  n = strlen(&s[1]);
  build(o, 1, n);
  int m;
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &a, &b);
    printf("%d\n", 2 * find(1, 1, n).tot);
  }
  return 0;
}
