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;
string p;
struct R {
  long long int open, close, len;
  R() { open = close = len = 0; }
};
R seg[4 * 1000005];
void combine(R a, R b, R &c) {
  long long int f = min(a.open, b.close);
  c.open = a.open + b.open - f;
  c.close = a.close + b.close - f;
  c.len = a.len + b.len + 2 * f;
}
void build(long long int node, long long int s, long long int e) {
  if (s == e) {
    if (p[s] == '(')
      seg[node].open++;
    else
      seg[node].close++;
    return;
  }
  build(2 * node, s, (s + e) / 2);
  build(2 * node + 1, (s + e) / 2 + 1, e);
  combine(seg[2 * node], seg[2 * node + 1], seg[node]);
}
R query(long long int node, long long int s, long long int e, long long int x,
        long long int y) {
  if (s > e or s > y or e < x) {
    R y;
    return y;
  }
  if (s >= x and e <= y) return seg[node];
  long long int mid = (s + e) / 2;
  R p1 = query(2 * node, s, mid, x, y);
  R p2 = query(2 * node + 1, mid + 1, e, x, y);
  R w;
  combine(p1, p2, w);
  return w;
}
int main() {
  long long int i, j, k, q, x, y;
  cin >> p;
  cin >> q;
  build(1, 0, (long long int)p.size() - 1);
  while (q--) {
    cin >> x >> y;
    x--, y--;
    cout << query(1, 0, (long long int)p.size() - 1, x, y).len << '\n';
  }
}
