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 + 5;
string a;
struct NODE {
  int o, c, ans;
} tree[5 * MAX], dummy;
NODE combine(NODE left, NODE right) {
  NODE p;
  int x = min(left.o, right.c);
  p.ans = (left.ans + right.ans) + x;
  p.o = (left.o + right.o) - x;
  p.c = (left.c + right.c) - x;
  return p;
}
void build(int node, int start, int end) {
  if (start == end) {
    tree[node].ans = 0;
    if (a[start] == '(')
      tree[node].o = 1, tree[node].c = 0;
    else
      tree[node].o = 0, tree[node].c = 1;
  } else {
    int mid = (start + end) >> 1;
    int left = node << 1;
    int right = left + 1;
    build(left, start, mid);
    build(right, mid + 1, end);
    tree[node] = combine(tree[left], tree[right]);
  }
}
NODE query(int node, int start, int end, int l, int r) {
  if (l <= start and end <= r) return tree[node];
  if (r < start or end < l) return dummy;
  int mid = (start + end) >> 1;
  int left = node << 1;
  int right = left + 1;
  NODE p1 = query(left, start, mid, l, r);
  NODE p2 = query(right, mid + 1, end, l, r);
  return combine(p1, p2);
}
int main(int argc, char* argv[]) {
  if (argc == 2 or argc == 3) freopen(argv[1], "r", stdin);
  if (argc == 3) freopen(argv[2], "w", stdout);
  int q, l, r;
  NODE p;
  cin >> a;
  dummy.ans = 0;
  dummy.o = 0;
  dummy.c = 0;
  build(1, 0, a.length() - 1);
  cin >> q;
  for (int i = 0; i < q; i++) {
    cin >> l >> r;
    l--, r--;
    p = query(1, 0, a.length() - 1, l, r);
    cout << 2 * p.ans << endl;
  }
  return 0;
}
