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;
const int N = 1e6 + 5;
int sum[N * 2], l[N * 2], r[N * 2], n;
void build() {
  int i, mini;
  for (i = n - 1; i > 0; --i) {
    mini = min(l[i << 1], r[i << 1 | 1]);
    l[i] = l[i << 1] - mini + l[i << 1 | 1],
    r[i] = r[i << 1 | 1] - mini + r[i << 1];
    sum[i] = (sum[i << 1] + sum[i << 1 | 1] + mini);
  }
}
struct r {
  int ret, L, R;
  void init() { ret = 0, L = 0, R = 0; }
} x, y;
int query(int a, int b) {
  x.init(), y.init();
  int mini;
  for (a += n, b += n; a < b; a >>= 1, b >>= 1) {
    if (a & 1) {
      x.ret += sum[a];
      x.ret += min(r[a], x.L);
      x.L -= min(r[a], x.L);
      x.L += l[a];
      ++a;
    }
    if (b & 1) {
      --b;
      mini = min(l[b], y.R);
      y.R -= mini;
      y.R += r[b];
      y.ret += mini + sum[b];
    }
  }
  mini = min(x.L, y.R);
  return 2 * (x.ret + y.ret + mini);
}
int main() {
  string s;
  cin >> s;
  n = s.size();
  int i, q, a, b;
  for (i = 0; i < n; ++i)
    if (s[i] == '(') {
      l[i + n] = 1;
    } else {
      r[i + n] = 1;
    }
  build();
  cin >> q;
  while (q--) {
    scanf("%d%d", &a, &b);
    --a;
    printf("%d\n", query(a, b));
  }
}
