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 TAM = 1000000 + 10;
const long long MOD = 1000000007LL;
struct nodo {
  int c, pos, neg;
};
struct nodo s[4 * TAM], nil;
int n;
string a;
struct nodo merge(struct nodo e1, struct nodo e2) {
  struct nodo aux;
  int nuevos = min(e1.pos, e2.neg);
  aux.c = e1.c + e2.c + nuevos;
  aux.pos = e1.pos + e2.pos - nuevos;
  aux.neg = e1.neg + e2.neg - nuevos;
  return aux;
}
void build(int p, int l, int r) {
  if (l == r) {
    s[p] = nil;
    if (a[l] == '(')
      s[p].pos++;
    else
      s[p].neg++;
    return;
  }
  int mid = (l + r) / 2;
  build(2 * p, l, mid);
  build(2 * p + 1, mid + 1, r);
  s[p] = merge(s[2 * p], s[2 * p + 1]);
}
struct nodo query(int p, int l, int r, int i, int j) {
  if (j < l || r < i) return nil;
  if (i <= l && r <= j) return s[p];
  int mid = (l + r) / 2;
  struct nodo q1, q2;
  q1 = query(2 * p, l, mid, i, j);
  q2 = query(2 * p + 1, mid + 1, r, i, j);
  return merge(q1, q2);
}
int main() {
  nil.c = 0;
  nil.pos = 0;
  nil.neg = 0;
  cin >> a;
  n = a.size();
  build(1, 0, n - 1);
  int q, a, b;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    scanf("%d%d", &a, &b);
    printf("%d\n", query(1, 0, n - 1, a - 1, b - 1).c * 2);
  }
  return 0;
}
