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 N = 1000005;
int a[N];
struct SegTree {
  int data;
};
SegTree tree[4 * N];
void build(int st_index, int low, int high) {
  if (low == high) {
    tree[st_index].data = a[low];
    return;
  }
  int mid = (low + high) / 2;
  int left = 2 * st_index + 1;
  int right = 2 * st_index + 2;
  build(left, low, mid);
  build(right, mid + 1, high);
  tree[st_index].data = min(tree[left].data, tree[right].data);
}
int get(int seg_start, int seg_end, int query_start, int query_end,
        int st_index) {
  if (seg_start >= query_start && seg_end <= query_end)
    return tree[st_index].data;
  if (seg_end < query_start || seg_start > query_end) return INT_MAX;
  int mid = (seg_start + seg_end) / 2;
  return min(get(seg_start, mid, query_start, query_end, 2 * st_index + 1),
             get(mid + 1, seg_end, query_start, query_end, 2 * st_index + 2));
}
int main() {
  string s;
  cin >> s;
  int n = s.length();
  a[0] = 0;
  for (int i = 0; i < n; i++) {
    a[i + 1] = a[i] + (s[i] == '(' ? 1 : -1);
  }
  build(0, 0, n);
  int m, x, y, ans, balance;
  cin >> m;
  for (int i = 0; i < m; i++) {
    cin >> x >> y;
    x--;
    balance = get(0, n, x, y, 0);
    int leftbal = a[x];
    int rightbal = a[y];
    int skipCloses = leftbal - balance;
    int skipOpens = rightbal - balance;
    printf("%d\n", y - x - skipCloses - skipOpens);
  }
  return 0;
}
