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;
struct segment_tree {
  int len, left, right;
};
segment_tree t[4 * 1000001], ans;
int n, m, cnt;
char s[1000001];
void bulid(int v, int tl, int tr);
;
void get_ans(int l, int r, int v, int tl, int tr);
void build(int v = 1, int tl = 0, int tr = cnt) {
  if (tl > tr) return;
  if (tl == tr) {
    if (s[tl] == '(')
      t[v].left = 1;
    else
      t[v].left = 0;
    t[v].right = 1 - t[v].left;
    return;
  }
  int tm = (tl + tr) / 2;
  build(2 * v, tl, tm);
  build(2 * v + 1, tm + 1, tr);
  t[v].len = t[2 * v].len + t[2 * v + 1].len +
             min(t[2 * v].left, t[2 * v + 1].right) * 2;
  t[v].left = t[2 * v].left + t[2 * v + 1].left -
              min(t[2 * v].left, t[2 * v + 1].right);
  t[v].right = t[2 * v].right + t[2 * v + 1].right -
               min(t[2 * v].left, t[2 * v + 1].right);
  return;
}
void get_ans(int l, int r, int v = 1, int tl = 0, int tr = cnt) {
  if (l > r) return;
  if (tl == l && tr == r) {
    ans.len += t[v].len + min(ans.left, t[v].right) * 2;
    ans.left += t[v].left - min(ans.left, t[v].right);
    ans.right += t[v].right - min(ans.left, t[v].right);
    return;
  }
  int tm = (tl + tr) / 2;
  get_ans(l, min(tm, r), 2 * v, tl, tm);
  get_ans(max(l, tm + 1), r, 2 * v + 1, tm + 1, tr);
  return;
}
int main() {
  cin.getline(s, 1000001);
  n = strlen(s);
  for (cnt = 1; cnt < n; cnt <<= 1)
    ;
  cnt--;
  build(1, 0, cnt);
  scanf("%i", &m);
  while (m > 1) {
    int l, r;
    scanf("%i %i", &l, &r);
    l--, r--;
    ans.len = ans.left = ans.right = 0;
    get_ans(l, r);
    printf("%i\n", ans.len);
    m--;
  }
  int l, r;
  scanf("%i %i", &l, &r);
  l--, r--;
  ans.len = ans.left = ans.right = 0;
  get_ans(l, r);
  printf("%i", ans.len);
  return 0;
}
