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 n, tre[4000015], a[1000005];
char str[1000005];
void Create(int l, int r, int x) {
  int m;
  if (l == r) {
    tre[x] = a[r];
    return;
  }
  m = (l + r) / 2;
  Create(l, m, x * 2);
  Create(m + 1, r, x * 2 + 1);
  tre[x] = min(tre[x * 2], tre[x * 2 + 1]);
}
int Query(int l, int r, int x, int c, int d) {
  int m, now;
  if (l >= c && r <= d) return tre[x];
  m = (l + r) / 2;
  now = 1000000;
  if (c <= m) now = min(now, Query(l, m, x * 2, c, d));
  if (d >= m + 1) now = min(now, Query(m + 1, r, x * 2 + 1, c, d));
  return now;
}
int main(void) {
  int x, y, T, i, bet;
  scanf("%s%d", str + 1, &T);
  n = strlen(str + 1);
  for (i = 1; i <= n; i++) a[i] = a[i - 1] + (str[i] == '(' ? 1 : -1);
  Create(1, n, 1);
  while (T--) {
    scanf("%d%d", &x, &y);
    bet = min(Query(1, n, 1, x, y), a[x - 1]) - a[x - 1];
    printf("%d\n", y - x + 1 + bet - max(a[y] - a[x - 1] - bet, 0));
  }
  return 0;
}
