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 open, closed, max_sec;
  node(int a = 0, int b = 0, int c = 0) {
    open = a;
    closed = b;
    max_sec = c;
  }
};
string A;
int from, to;
class SegmentTree {
 public:
  vector<node> ST;
  SegmentTree(int num) { ST.resize(4 * num); }
  node join(node left, node right) {
    int mn = min(left.open, right.closed);
    int a = left.open + right.open - mn;
    int b = left.closed + right.closed - mn;
    int c = left.max_sec + right.max_sec + 2 * mn;
    return node(a, b, c);
  }
  void bulid(int p, int start, int end) {
    if (start == end) {
      ST[p].open = (A[start] == '(');
      ST[p].closed = (A[start] == ')');
      return;
    }
    int mid = (start + end) / 2;
    bulid(2 * p, start, mid);
    bulid(2 * p + 1, mid + 1, end);
    ST[p] = join(ST[2 * p], ST[2 * p + 1]);
  }
  node get(int p, int start, int end) {
    if (start > end || end < from || start > to) return node();
    int mid = (start + end) / 2;
    if (from <= start && to >= end) return ST[p];
    node a = get(2 * p, start, mid);
    node b = get(2 * p + 1, mid + 1, end);
    return join(a, b);
  }
};
int main() {
  cin >> A;
  int m, a, b;
  SegmentTree ST(A.size());
  ST.bulid(1, 0, A.size() - 1);
  cin >> m;
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &from, &to);
    from--, to--;
    printf("%d\n", ST.get(1, 0, A.size() - 1).max_sec);
  }
}
