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 Segment {
  int opening, closing, matching;
  void set_segment(int x, int y, int z) {
    opening = x, closing = y, matching = z;
  }
} T[2097160];
char s[1000006];
Segment INV;
Segment combine_segments(Segment &lhs, Segment &rhs) {
  int newMatch = min(lhs.opening, rhs.closing);
  Segment seg;
  seg.set_segment(lhs.opening - newMatch + rhs.opening,
                  lhs.closing + rhs.closing - newMatch,
                  lhs.matching + rhs.matching + newMatch);
  return seg;
}
void build_segment(int v, int l, int r) {
  if (l == r)
    T[v].set_segment(s[l] == '(', s[l] == ')', 0);
  else {
    build_segment(((v) << 1), l, (l + r) / 2),
        build_segment((((v) << 1) | 1), (l + r) / 2 + 1, r);
    T[v] = combine_segments(T[((v) << 1)], T[(((v) << 1) | 1)]);
  }
}
void build_tree(int n) { build_segment(1, 0, n - 1); }
Segment query(int v, int l, int r, int &i, int &j) {
  if (r < i || l > j) return INV;
  if (l >= i && r <= j) return T[v];
  Segment left = query(((v) << 1), l, (l + r) / 2, i, j),
          right = query((((v) << 1) | 1), (l + r) / 2 + 1, r, i, j);
  if (left.matching == -1) return right;
  if (right.matching == -1) return left;
  return combine_segments(left, right);
}
void solve(int q) {
  int n = strlen(s), l, r;
  INV.matching = -1;
  build_tree(n);
  while (q--) {
    scanf("%d %d", &l, &r);
    printf("%d\n", 2 * query(1, 0, n - 1, --l, --r).matching);
  }
}
int main() {
  int q;
  scanf("%s %d", s, &q);
  solve(q);
  return 0;
}
