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 N = 1e6;
struct tree {
  int mx, turnLeft, turnRight;
};
string a;
int n, m;
tree t[4 * N], e;
tree combine(tree t1, tree t2) {
  tree res;
  int add = min(t1.turnRight, t2.turnLeft);
  res.mx = t1.mx + t2.mx + add;
  res.turnRight = t1.turnRight + t2.turnRight - add;
  res.turnLeft = t1.turnLeft + t2.turnLeft - add;
  return res;
}
void build(int v, int tl, int tr) {
  if (tl == tr) {
    if (a[tl] == '(')
      t[v].turnRight = 1;
    else
      t[v].turnLeft = 1;
  } else {
    int tm = (tl + tr) >> 1;
    build(v * 2 + 1, tl, tm);
    build(v * 2 + 2, tm + 1, tr);
    t[v] = combine(t[v * 2 + 1], t[v * 2 + 2]);
  }
}
tree get(int v, int tl, int tr, int l, int r) {
  if (l > r) return e;
  if (tl == l && tr == r) return t[v];
  int tm = (tl + tr) >> 1;
  return combine(get(v * 2 + 1, tl, tm, l, min(tm, r)),
                 get(v * 2 + 2, tm + 1, tr, max(l, tm + 1), r));
}
int main() {
  cin >> a;
  cin >> m;
  n = a.size();
  build(0, 0, n - 1);
  for (int i = 0; i < m; i++) {
    int l, r;
    cin >> l >> r;
    int ans = get(0, 0, n - 1, l - 1, r - 1).mx;
    cout << ans * 2 << endl;
  }
  return 0;
}
