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 double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int N = 1e6 + 9;
struct node {
  int a, b, c;
  node(int first = 0, int second = 0, int z = 0) : a(first), b(second), c(z) {}
} tree[N * 4];
string s;
int n, len, n1;
void build() {
  n1 = 1;
  while (n1 < len) n1 *= 2;
  for (int i = 0; i < len; i++) {
    if (s[i] == '(')
      tree[i + n1 - 1].a++;
    else
      tree[i + n1 - 1].b++;
  }
  for (int i = n1 - 2; i >= 0; i--) {
    node p1 = tree[2 * i + 1], p2 = tree[2 * i + 2];
    int t = min(p1.a, p2.b);
    p1.a -= t, p2.b -= t;
    tree[i].c = p1.c + p2.c + t;
    tree[i].a = p1.a + p2.a;
    tree[i].b = p1.b + p2.b;
  }
}
node query(int node, int start, int end, int l, int r) {
  struct node ans;
  if (r < start || l > end) return ans;
  if (l <= start && end <= r) return tree[node];
  int mid = (start + end) / 2;
  struct node p1 = query(node * 2 + 1, start, mid, l, r);
  struct node p2 = query(node * 2 + 2, mid + 1, end, l, r);
  int t = min(p1.a, p2.b);
  p1.a -= t, p2.b -= t;
  ans.c = p1.c + p2.c + t;
  ans.a = p1.a + p2.a;
  ans.b = p1.b + p2.b;
  return ans;
}
int main(void) {
  cin >> s >> n;
  len = s.length();
  build();
  for (int i = 0; i < n; i++) {
    int l, r;
    cin >> l >> r;
    printf("%d\n", 2 * query(0, 0, n1 - 1, l - 1, r - 1).c);
  }
  return 0;
}
