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;
class segtree {
 public:
  int val[(1 << 20) * 2];
  int le[(1 << 20) * 2];
  int ri[(1 << 20) * 2];
  void update(int pos, int x) {
    val[pos + (1 << 20) - 1] = 0;
    le[pos + (1 << 20) - 1] = 1 - x;
    ri[pos + (1 << 20) - 1] = x;
  }
  void up() {
    for (int i = (1 << 20) - 2; i >= 0; i--) {
      int add = min(le[i * 2 + 1], ri[i * 2 + 2]);
      val[i] = val[i * 2 + 1] + val[i * 2 + 2] + add;
      le[i] = le[i * 2 + 1] + le[i * 2 + 2] - add;
      ri[i] = ri[i * 2 + 1] + ri[i * 2 + 2] - add;
    }
  }
  pair<int, pair<int, int> > query(int a, int b, int k, int l, int r) {
    pair<int, pair<int, int> > p = make_pair(0, make_pair(0, 0));
    if (r < a || b < l) return p;
    if (a <= l && r <= b) {
      return make_pair(val[k], make_pair(le[k], ri[k]));
    }
    pair<int, pair<int, int> > x = query(a, b, k * 2 + 1, l, (l + r) / 2);
    pair<int, pair<int, int> > y = query(a, b, k * 2 + 2, (l + r) / 2 + 1, r);
    int z = min(x.second.first, y.second.second);
    return make_pair(x.first + y.first + z,
                     make_pair(x.second.first + y.second.first - z,
                               x.second.second + y.second.second - z));
  }
} seg;
char ss[1000005];
int main() {
  scanf("%s", &ss);
  int n = strlen(ss);
  for (int i = 0; i < n; i++) {
    seg.update(i, (ss[i] == ')'));
  }
  seg.up();
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    a--;
    b--;
    printf("%d\n", seg.query(a, b, 0, 0, (1 << 20) - 1).first * 2);
  }
}
