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;
using namespace std;
struct node {
  int o, c, s;
};
node t[5000000];
node combine(node a, node b) {
  node ans;
  ans.s = a.s + b.s;
  int z = min(a.o, b.c);
  ans.o = a.o + b.o - z;
  ans.c = a.c + b.c - z;
  ans.s = ans.s + z + z;
  return ans;
}
void build(string &v, int i, int tl, int tr) {
  if (tl == tr) {
    node tt;
    if (v[tl] == ')') {
      tt.o = 0;
      tt.c = 1;
      tt.s = 0;
    } else {
      tt.o = 1;
      tt.c = 0;
      tt.s = 0;
    }
    t[i] = tt;
  } else {
    int tm = (tl + tr) / 2;
    build(v, 2 * i, tl, tm);
    build(v, 2 * i + 1, tm + 1, tr);
    t[i] = combine(t[2 * i], t[2 * i + 1]);
  }
}
node query(int i, int tl, int tr, int l, int r) {
  if (l > r) {
    node tt;
    tt.o = 0;
    tt.c = 0;
    tt.s = 0;
    return tt;
  }
  if (tl == l && tr == r) {
    return t[i];
  }
  int tm = (tl + tr) / 2;
  return combine(query(2 * i, tl, tm, l, min(tm, r)),
                 query(2 * i + 1, tm + 1, tr, max(l, tm + 1), r));
}
int main() {
  string s;
  cin >> s;
  int n = s.size();
  build(s, 1, 0, n - 1);
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    cout << query(1, 0, n - 1, a, b).s << endl;
  }
}
