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 l, r, ans;
  node() : l(0), r(0), ans(0) {}
  node(int a, int b, int c) : l(a), r(b), ans(c) {}
  void init() { l = 0, r = 0, ans = 0; }
} box[1111111 << 2];
string s;
int len;
inline void Merge(node& p, node& q, node& r) {
  int tmp = min(q.l, r.r);
  p.l = q.l + r.l - tmp;
  p.r = q.r + r.r - tmp;
  p.ans = q.ans + r.ans + tmp;
}
void build(int n, int b, int e) {
  int mid = b + (e - b) / 2;
  int lc = n << 1;
  int rc = lc + 1;
  if (b == e) {
    box[n].init();
    if (s.at(b) == '(')
      box[n].l = 1;
    else
      box[n].r = 1;
    return;
  }
  build(lc, b, mid);
  build(rc, mid + 1, e);
  Merge(box[n], box[lc], box[rc]);
}
int m, lp, rp;
node query(int n, int b, int e) {
  int mid = b + (e - b) / 2;
  int lc = n << 1;
  int rc = lc + 1;
  if (lp <= b && e <= rp) return box[n];
  if (e < lp || rp < b) return node(0, 0, 0);
  node afsana = query(lc, b, mid);
  node islam = query(rc, mid + 1, e);
  node a;
  a.init();
  Merge(a, afsana, islam);
  return a;
}
int main() {
  cin >> s;
  len = s.length();
  build(1, 0, len - 1);
  cin >> m;
  while (m--) {
    cin >> lp >> rp;
    lp--, rp--;
    cout << 2 * query(1, 0, len - 1).ans << "\n";
  }
  return 0;
}
