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>
#pragma comment(linker, "/stack:30000000")
using namespace std;
using namespace std;
struct Node {
  int open, close, ans;
  Node() {}
  Node(int open, int close, int ans) {
    open = open;
    close = close;
    ans = ans;
  }
};
Node X[3 * 1000010];
string A;
Node Invalid(-1, -1, -1);
void build(int i, int s, int f) {
  if (s == f) {
    X[i].ans = X[i].open = X[i].close = 0;
    if (A[s] == '(')
      X[i].open = 1;
    else
      X[i].close = 1;
    return;
  }
  int mid = (s + f) / 2;
  build(2 * i, s, mid);
  build(2 * i + 1, mid + 1, f);
  int temp = min(X[2 * i].open, X[2 * i + 1].close);
  X[i].ans = X[2 * i].ans + X[2 * i + 1].ans + temp;
  X[i].open = X[2 * i].open + X[2 * i + 1].open - temp;
  X[i].close = X[2 * i].close + X[2 * i + 1].close - temp;
}
pair<pair<int, int>, int> query(int i, int s, int f, int x, int y) {
  if (x > f || y < s) return make_pair(make_pair(-1, -1), -1);
  if (s >= x && f <= y)
    return make_pair(make_pair(X[i].open, X[i].close), X[i].ans);
  int mid = (s + f) / 2;
  pair<pair<int, int>, int> l = query(2 * i, s, mid, x, y);
  pair<pair<int, int>, int> r = query(2 * i + 1, mid + 1, f, x, y);
  int temp = min(l.first.first, r.first.second);
  if (l.second == -1)
    return r;
  else if (r.second == -1)
    return l;
  else
    return make_pair(make_pair(l.first.first + r.first.first - temp,
                               l.first.second + r.first.second - temp),
                     l.second + r.second + temp);
}
int main() {
  cin >> A;
  int m, x, y, n = A.size();
  build(1, 0, n - 1);
  cin >> m;
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &x, &y);
    x--, y--;
    printf("%d\n", 2 * query(1, 0, n - 1, x, y).second);
  }
}
