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 MAX = 1e6 + 1;
struct node {
  int ans = 0;
  int o = 0;
  int c = 0;
  node() : ans(0), o(0), c(0) {}
  node(int x) : ans(0), o((x == 1 ? 1 : 0)), c((x == 1 ? 0 : 1)) {}
};
node t[2 * MAX];
int n;
node combine(node x, node y) {
  node cur;
  int t = min(x.o, y.c);
  cur.ans = t + x.ans + y.ans;
  cur.o = x.o + y.o - t;
  cur.c = x.c + y.c - t;
  return cur;
}
int query(int l, int r) {
  node resl, resr;
  for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
    if (l & 1) resl = combine(resl, t[l++]);
    if (r & 1) resr = combine(t[--r], resr);
  }
  return combine(resl, resr).ans;
}
int main() {
  string s;
  cin >> s;
  n = s.length();
  for (int i = 0; i < n; i++) {
    int ch = (s[i] == '(' ? 1 : 0);
    t[i + n] = node(ch);
  }
  for (int i = n - 1; i > 0; i--) {
    int x = i << 1;
    int y = i << 1 | 1;
    if (x > y) {
      t[i] = combine(t[y], t[x]);
    } else {
      t[i] = combine(t[x], t[y]);
    }
  }
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    printf("%d\n", query(l - 1, r) * 2);
  }
  return 0;
}
