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;
char s[1000001];
int ans[4 * 1000001], st[4 * 1000001], cl[4 * 1000001];
void init(int l, int r, int id) {
  if (l == r) {
    if (s[l] == '(')
      st[id] = 1;
    else
      cl[id] = 1;
    return;
  }
  int mid = (l + r) / 2, lf = id << 1, ri = lf + 1;
  init(l, mid, lf), init(mid + 1, r, ri);
  int mx = min(st[lf], cl[ri]);
  ans[id] = ans[lf] + ans[ri] + mx;
  st[id] = st[lf] + st[ri] - mx;
  cl[id] = cl[lf] + cl[ri] - mx;
}
struct node {
  int st, cl, bst;
  node(int a, int b, int c) {
    st = a;
    cl = b;
    bst = c;
  }
  node();
};
node query(int l, int r, int id, int x, int y) {
  if (l > y || r < x) return node(0, 0, 0);
  if (l >= x && r <= y) return node(st[id], cl[id], ans[id]);
  int mid = (l + r) / 2, lf = id << 1, ri = lf + 1;
  node a = query(l, mid, lf, x, y), b = query(mid + 1, r, ri, x, y);
  node ss(0, 0, 0);
  int mx = min(a.st, b.cl);
  ss.st = a.st + b.st - mx;
  ss.cl = a.cl + b.cl - mx;
  ss.bst = a.bst + b.bst + mx;
  return ss;
}
int main() {
  scanf("%s", s);
  int n = strlen(s);
  init(0, n - 1, 1);
  int q, x, y;
  scanf("%d", &q);
  while (q--) {
    scanf("%d %d", &x, &y);
    node res = query(0, n - 1, 1, x - 1, y - 1);
    printf("%d\n", res.bst * 2);
  }
  return 0;
}
