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;
int l, r, m;
string s;
struct yo {
  int a, b, c;
};
yo make_yo(int c, int b, int a) {
  yo dich;
  dich.a = a;
  dich.b = b;
  dich.c = c;
  return dich;
}
int a[4 * 1000000];
int b[4 * 1000000];
int c[4 * 1000000];
void build(int t, int tl, int tr) {
  if (tl == tr) {
    c[t] = 0;
    if (s[tl - 1] == '(') {
      a[t] = 1;
      b[t] = 0;
    } else {
      b[t] = 1;
      a[t] = 0;
    }
    return;
  }
  int tm = (tl + tr) / 2;
  build(t * 2, tl, tm);
  build(t * 2 + 1, tm + 1, tr);
  c[t] = c[t * 2] + c[t * 2 + 1] + 2 * min(a[t * 2], b[t * 2 + 1]);
  a[t] = a[t * 2] + a[t * 2 + 1] - min(a[t * 2], b[t * 2 + 1]);
  b[t] = b[t * 2] + b[t * 2 + 1] - min(a[t * 2], b[t * 2 + 1]);
}
yo getls(int t, int l, int r, int tl, int tr) {
  if (l > r) return make_yo(0, 0, 0);
  if (tl == l && tr == r) return make_yo(c[t], b[t], a[t]);
  int tm = (tl + tr) / 2;
  yo p1 = getls(t * 2, l, min(tm, r), tl, tm);
  yo p2 = getls(t * 2 + 1, max(tm + 1, l), r, tm + 1, tr);
  yo newp;
  newp.c = p1.c + p2.c + 2 * min(p1.a, p2.b);
  newp.a = p1.a + p2.a - min(p1.a, p2.b);
  newp.b = p1.b + p2.b - min(p1.a, p2.b);
  return newp;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> s >> m;
  build(1, 1, s.length());
  for (int i = 1; i <= m; i++) {
    cin >> l >> r;
    cout << getls(1, l, r, 1, s.length()).c << '\n';
  }
}
