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;
const long long mod = 258280327;
int sum[N << 2], a[N << 2];
char s[N];
void pushup(int rt) { sum[rt] = min(sum[rt << 1], sum[rt << 1 | 1]); }
void build(int l, int r, int rt) {
  if (l == r) {
    sum[rt] = a[l];
    return;
  }
  int m = (l + r) >> 1;
  build(l, m, rt << 1);
  build(m + 1, r, rt << 1 | 1);
  pushup(rt);
}
int query(int L, int R, int l, int r, int rt) {
  if (L > r || R < l) return 0x3f3f3f3f;
  if (L <= l && R >= r) return sum[rt];
  int m = (l + r) >> 1;
  return min(query(L, R, l, m, rt << 1), query(L, R, m + 1, r, rt << 1 | 1));
}
int main() {
  scanf("%s", s);
  int n = strlen(s);
  for (int i = 0; i < n; i++) a[i + 1] = a[i] + (s[i] == '(' ? 1 : -1);
  build(0, n, 1);
  int m;
  scanf("%d", &m);
  while (m--) {
    int l, r;
    scanf("%d%d", &l, &r);
    l--;
    int d = query(l, r, 0, n, 1);
    printf("%d\n",
           r - l - (query(l, l, 0, n, 1) - d) - (query(r, r, 0, n, 1) - d));
  }
  return 0;
}
