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;
long long INF = 1e10;
string s;
struct node {
  int ans;
  int o;
  int c;
  node(int a = 0, int b = 0, int d = 0) {
    ans = a;
    o = b;
    c = d;
  }
};
node t[4004000];
node merge(node a, node b) {
  node c(0, 0, 0);
  c.ans = a.ans + b.ans;
  c.o = a.o + b.o;
  c.c = a.c + b.c;
  if (a.o > b.c) {
    c.ans += 2 * b.c;
    c.o -= b.c;
    c.c -= b.c;
  } else {
    c.ans += 2 * a.o;
    c.o -= a.o;
    c.c -= a.o;
  }
  return c;
}
void build(long long index, long long l, long long r) {
  if (l == r) {
    if (s[l] == '(')
      t[index] = node(0, 1, 0);
    else
      t[index] = node(0, 0, 1);
    return;
  }
  long long mid = (l + r) / 2;
  build(2 * index, l, mid);
  build(2 * index + 1, mid + 1, r);
  t[index] = merge(t[2 * index], t[2 * index + 1]);
}
node query(long long index, long long l, long long r, long long lq,
           long long rq) {
  if (r < lq || l > rq) return node(0, 0, 0);
  if (l >= lq && r <= rq) return t[index];
  long long mid = (l + r) / 2;
  return merge(query(2 * index, l, mid, lq, rq),
               query(2 * index + 1, mid + 1, r, lq, rq));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> s;
  long long n = s.length();
  long long q;
  cin >> q;
  build(1, 0, n - 1);
  for (long long i = 0; i <= q - 1; ++i) {
    int x, y;
    cin >> x;
    if (x == 0)
      return 0;
    else {
      cin >> y;
      x--;
      y--;
      node z;
      z = query(1, 0, n - 1, x, y);
      cout << z.ans << "\n";
    }
  }
}
