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 long long inf = 1e9 + 7;
struct tr {
  long long op;
  long long cl;
  long long res;
} tree[4 * 1000000 + 5];
string second;
long long n, q;
inline tr merge(tr p1, tr p2) {
  int temp = min(p1.op, p2.cl);
  tr ans;
  ans.op = p1.op + p2.op - temp;
  ans.cl = p1.cl + p2.cl - temp;
  ans.res = p1.res + p2.res + temp;
  return ans;
}
void build(int node, int l, int r) {
  if (l == r) {
    tree[node].res = 0;
    tree[node].op = (second[l] == '(');
    tree[node].cl = (second[l] == ')');
    return;
  }
  int mid = (l + r) / 2;
  build(2 * node, l, mid);
  build(2 * node + 1, mid + 1, r);
  tree[node] = merge(tree[2 * node], tree[2 * node + 1]);
}
tr query(int node, int l, int r, int lq, int rq) {
  if (lq == l && r == rq) {
    return tree[node];
  }
  int mid = (l + r) / 2;
  tr var;
  var.op = 0;
  var.cl = 0;
  var.res = 0;
  if (lq <= mid) var = merge(var, query(node * 2, l, mid, lq, min(mid, rq)));
  if (mid < rq)
    var = merge(var, query(node * 2 + 1, mid + 1, r, max(mid + 1, lq), rq));
  return var;
  return var;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> second;
  int n = second.size();
  second = "0" + second;
  build(1, 1, n);
  cin >> q;
  int x, y;
  while (q--) {
    cin >> x >> y;
    int ans = 2 * (query(1, 1, n, x, y).res);
    cout << ans << endl;
  }
}
