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 Node {
  int o;
  int c;
  int m;
  Node* left;
  Node* right;
  Node() : o(0), c(0), m(0), left(0), right(0) {}
};
string s;
Node nodes[4000005];
int K;
int N;
Node* new_node() { return &nodes[K++]; }
Node combine(Node* left_node, Node* right_node) {
  Node ret;
  int matched = min(left_node->o, right_node->c);
  ret.m = left_node->m + right_node->m + matched;
  ret.o = left_node->o + right_node->o - matched;
  ret.c = left_node->c + right_node->c - matched;
  return ret;
}
void build(Node* node, int l = 0, int r = N - 1) {
  if (l == r) {
    node->o = s[l] == '(';
    node->c = s[l] == ')';
    return;
  }
  int m = (l + r) / 2;
  node->left = new_node();
  node->right = new_node();
  build(node->left, l, m);
  build(node->right, m + 1, r);
  Node k = combine(node->left, node->right);
  node->o = k.o;
  node->c = k.c;
  node->m = k.m;
}
Node query(int x, int y, Node* node, int l = 0, int r = N - 1) {
  if (y < l || r < x) return Node();
  if (x <= l && r <= y) return *node;
  int m = (l + r) / 2;
  Node left = query(x, y, node->left, l, m);
  Node right = query(x, y, node->right, m + 1, r);
  return combine(&left, &right);
}
int main() {
  cin >> s;
  N = s.size();
  int m;
  cin >> m;
  Node* root = new_node();
  build(root);
  for (int i = 0; i < m; ++i) {
    int x, y;
    cin >> x >> y;
    cout << query(x - 1, y - 1, root).m * 2 << '\n';
  }
}
