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 int mod = 1e9 + 7;
const double eps = 1e-18;
const double PI = atan(1.0);
const int NMAX = 1234567;
int tree[NMAX * 2];
int open[NMAX * 2];
int close[NMAX * 2];
string s;
int n;
void build(int l, int r, int node) {
  if (l == r) {
    if (s[l] == '(')
      tree[node] = 0, open[node] = 1, close[node] = 0;
    else
      tree[node] = 0, close[node] = 1, open[node] = 0;
  } else {
    int mid = (l + r) >> 1;
    build(l, mid, node << 1);
    build(mid + 1, r, node << 1 | 1);
    int tmp = min(open[node << 1], close[node << 1 | 1]);
    tree[node] = tree[node << 1] + tree[node << 1 | 1] + tmp;
    open[node] = open[node << 1] + open[node << 1 | 1] - tmp;
    close[node] = close[node << 1] + close[node << 1 | 1] - tmp;
  }
}
struct Node {
  int o, c, t;
  Node(int t, int o, int c) {
    this->t = t;
    this->c = c;
    this->o = o;
  }
};
Node query(int l, int r, int ll, int rr, int node) {
  if (l > rr || r < ll) return Node(0, 0, 0);
  if (l >= ll && r <= rr) return Node(tree[node], open[node], close[node]);
  int mid = (l + r) >> 1;
  Node node1 = query(l, mid, ll, rr, node << 1),
       node2 = query(mid + 1, r, ll, rr, node << 1 | 1);
  int tmp = min(node1.o, node2.c);
  return Node(node1.t + node2.t + tmp, node1.o + node2.o - tmp,
              node1.c + node2.c - tmp);
}
int main() {
  ios::sync_with_stdio(0);
  ;
  cin >> s;
  s.insert(s.begin(), '.');
  n = s.size();
  build(1, n, 1);
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int l, r;
    cin >> l >> r;
    cout << query(1, n, l, r, 1).t * 2 << "\n";
  }
}
