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 node {
  int ans, open, close;
};
int n;
node t[4 * 1000005];
node combine(node A, node B) {
  node temp;
  temp.open = A.open + B.open;
  temp.close = A.close + B.close;
  temp.ans =
      A.ans + B.ans + 2 * (min(A.open - (A.ans / 2), B.close - (B.ans / 2)));
  return temp;
}
void build(string &s, int v, int tl, int tr) {
  if (tl == tr) {
    t[v].ans = 0;
    if (s[tl] == '(')
      t[v].open = 1;
    else
      t[v].close = 1;
  } else {
    int tm = (tl + tr) / 2;
    build(s, v * 2, tl, tm);
    build(s, v * 2 + 1, tm + 1, tr);
    t[v] = combine(t[v * 2], t[v * 2 + 1]);
  }
}
node query(int v, int tl, int tr, int l, int r) {
  if (l > r) {
    node temp;
    return temp;
  }
  if (l == tl && r == tr) {
    return t[v];
  }
  int tm = (tl + tr) / 2;
  return combine(query(v * 2, tl, tm, l, min(r, tm)),
                 query(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r));
}
int main() {
  string s;
  cin >> s;
  int n = s.size();
  build(s, 1, 0, n - 1);
  int m;
  scanf("%d", &m);
  ;
  for (int i = 0; i < m; i++) {
    int l, r;
    scanf("%d", &l);
    ;
    scanf("%d", &r);
    ;
    printf("%d\n", query(1, 0, n - 1, l - 1, r - 1).ans);
    ;
  }
}
