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 Node {
 public:
  int o, c, otp;
  Node(){};
  Node(int _o, int _c, int _otp) {
    o = _o;
    c = _c;
    otp = _otp;
  }
};
Node merge(Node x, Node y) {
  Node ans;
  int tmp = min(x.o, y.c);
  ans.otp = x.otp + y.otp + 2 * tmp;
  ans.c = x.c + y.c - tmp;
  ans.o = x.o + y.o - tmp;
  return ans;
}
void makeTree(int *a, Node *T, int id, int l, int h) {
  if (l == h) {
    if (a[l] == 0) {
      T[id] = Node(1, 0, 0);
    } else
      T[id] = Node(0, 1, 0);
    return;
  }
  int m = (h + l) / 2;
  makeTree(a, T, 2 * id + 1, l, m);
  makeTree(a, T, 2 * id + 2, m + 1, h);
  T[id] = merge(T[2 * id + 1], T[2 * id + 2]);
}
void change(int *a, string s, int N) {
  for (int i = 0; i < N; i++) {
    if (s[i] == '(')
      a[i] = 0;
    else
      a[i] = 1;
  }
}
int *query(int *a, int l, int r) {
  int *ans = new int[r - l + 1];
  for (int i = 0; i < r - l + 1; i++) {
    ans[i] = a[l + i];
  }
  return ans;
}
Node getMax(Node *T, int id, int l, int h, int u, int v) {
  if (u > h || v < l) return Node(0, 0, 0);
  if (l >= u && h <= v) {
    return T[id];
  }
  int m = (h + l) / 2;
  Node leftMax = getMax(T, id * 2 + 1, l, m, u, v);
  Node rightMax = getMax(T, id * 2 + 2, m + 1, h, u, v);
  return merge(leftMax, rightMax);
}
int main() {
  int N, Q;
  string input;
  cin >> input;
  N = input.length();
  int *a = new int[N];
  change(a, input, N);
  Node *T = new Node[4 * N];
  makeTree(a, T, 0, 0, N - 1);
  cin >> Q;
  for (int i = 0; i < Q; i++) {
    int l, r;
    cin >> l >> r;
    cout << getMax(T, 0, 0, N - 1, l - 1, r - 1).otp << endl;
  }
  return 0;
}
