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;
stack<pair<char, int> > st;
int d, l, r, v[1000001];
string ss;
struct cell {
  int x = 0, l = 0, r = 0;
};
cell seg[4000001];
void build(int i, int s, int e) {
  if (s == e) {
    if (ss[s - 1] == ')')
      seg[i].l = 1;
    else
      seg[i].r = 1;
    return;
  }
  build(i * 2, s, (s + e) / 2);
  build(i * 2 + 1, (s + e) / 2 + 1, e);
  cell &t = seg[i];
  cell q = seg[i * 2];
  cell w = seg[i * 2 + 1];
  t.x = q.x + w.x;
  int d = min(q.r, w.l);
  q.r -= d;
  w.l -= d;
  t.x += d;
  t.l = q.l + w.l;
  t.r = q.r + w.r;
}
cell get(int i, int s, int e) {
  if (e < l || s > r) {
    cell t;
    return t;
  }
  if (s >= l && e <= r) return seg[i];
  cell q = get(i * 2, s, (s + e) / 2);
  cell w = get(i * 2 + 1, (s + e) / 2 + 1, e);
  cell t;
  t.x = q.x + w.x;
  int d = min(q.r, w.l);
  q.r -= d;
  w.l -= d;
  t.x += d;
  t.l = q.l + w.l;
  t.r = q.r + w.r;
  return t;
}
int main() {
  cin >> ss;
  build(1, 1, ss.size());
  int n;
  cin >> n;
  while (n--) {
    scanf("%d%d", &l, &r);
    printf("%d\n", get(1, 1, ss.size()).x * 2);
  }
}
