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 tree_node {
  int o;
  int c;
  int b;
};
tree_node tree[1 << 21];
string s1;
tree_node mergee(tree_node a, tree_node b) {
  tree_node c;
  int x = min(a.o, b.c);
  c.b = a.b + b.b + x;
  c.o = a.o + b.o - x;
  c.c = a.c + b.c - x;
  return c;
}
void build(int node, int start, int end) {
  if (start == end) {
    if (s1[start] == ')')
      tree[node].c++;
    else
      tree[node].o++;
  } else {
    int mid = (start + end) / 2;
    build(2 * node, start, mid);
    build(2 * node + 1, mid + 1, end);
    tree[node] = mergee(tree[2 * node], tree[2 * node + 1]);
  }
}
tree_node create() {
  tree_node a;
  a.o = a.c = a.b = 0;
  return a;
}
tree_node query(int node, int start, int end, int l, int r) {
  if (r < start or end < l) {
    return create();
  }
  if (l <= start and end <= r) {
    return tree[node];
  }
  int mid = (start + end) / 2;
  tree_node p1 = query(2 * node, start, mid, l, r);
  tree_node p2 = query(2 * node + 1, mid + 1, end, l, r);
  return mergee(p1, p2);
}
int main() {
  cin >> s1;
  s1 = '#' + s1;
  int q;
  build(1, 1, s1.size());
  scanf("%d", &q);
  while (q--) {
    int l, r;
    scanf("%d %d", &l, &r);
    printf("%d\n", query(1, 1, s1.size(), l, r).b * 2);
  }
  return 0;
}
