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, o, c;
};
string a;
node tree[4 * 1000000];
void build(int s, int e, int i) {
  if (s == e) {
    if (a[s] == '(') {
      tree[i].o = 1;
      tree[i].c = tree[i].ans = 0;
    } else {
      tree[i].c = 1;
      tree[i].ans = tree[i].o = 0;
    }
    return;
  }
  int mid = (s + e) / 2;
  build(s, mid, 2 * i + 1);
  build(mid + 1, e, 2 * i + 2);
  int t = min(tree[2 * i + 1].o, tree[2 * i + 2].c);
  tree[i].ans = tree[2 * i + 1].ans + tree[2 * i + 2].ans + t;
  tree[i].o = tree[2 * i + 1].o + tree[2 * i + 2].o - t;
  tree[i].c = tree[2 * i + 1].c + tree[2 * i + 2].c - t;
}
node query(int s, int e, int i, int qs, int qe) {
  node n;
  n.o = n.c = 0;
  n.ans = 0;
  if (e < s || e < qs || s > qe) return n;
  if (s >= qs && e <= qe) {
    return tree[i];
  }
  int mid = (s + e) / 2;
  node p = query(s, mid, 2 * i + 1, qs, qe);
  node q = query(mid + 1, e, 2 * i + 2, qs, qe);
  int t = min(p.o, q.c);
  n.ans = p.ans + q.ans + t;
  n.o = p.o + q.o - t;
  n.c = p.c + q.c - t;
  return n;
}
int main() {
  int m, l, r;
  cin >> a;
  int n = a.length();
  build(0, n - 1, 0);
  cin >> m;
  while (m--) {
    cin >> l >> r;
    cout << query(0, n - 1, 0, l - 1, r - 1).ans * 2 << endl;
  }
  return 0;
}
