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 long long infl = 1e18 + 5;
struct tri {
  int a;
  int b;
  int c;
};
long long int m, n, q, x, y, k, mx = -1, mn = infl, f, val, sz, sm, cnt, ans,
                                t = 1, i, j, ind = -1;
string s;
std::vector<tri> tree(4000004);
void build(int node, int start, int end) {
  if (start == end) {
    if (s[start] == '(') {
      tree[node].a = 1;
      tree[node].b = 0;
      tree[node].c = 0;
    } else {
      tree[node].a = 0;
      tree[node].b = 0;
      tree[node].c = 1;
    }
    return;
  }
  int mid = (start + end) / 2;
  build(2 * node, start, mid);
  build(2 * node + 1, mid + 1, end);
  int mn = min(tree[2 * node].a, tree[2 * node + 1].c);
  tree[node].a = tree[2 * node].a + tree[2 * node + 1].a - mn;
  tree[node].c = tree[2 * node].c + tree[2 * node + 1].c - mn;
  tree[node].b = tree[2 * node].b + tree[2 * node + 1].b + mn;
}
tri query(int node, int start, int end, int l, int r) {
  tri t;
  t.a = 0;
  t.b = 0;
  t.c = 0;
  if (l > end || r < start) return t;
  if (start >= l && end <= r) {
    return tree[node];
  }
  int mid = (start + end) / 2;
  tri p1 = query(2 * node, start, mid, l, r);
  tri p2 = query(2 * node + 1, mid + 1, end, l, r);
  int mn = min(p1.a, p2.c);
  t.a = p1.a + p2.a - mn;
  t.c = p1.c + p2.c - mn;
  t.b = p1.b + p2.b + mn;
  return t;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  if (fopen("inp.txt", "r")) {
    freopen("myfile.txt", "w", stdout);
    freopen("inp.txt", "r", stdin);
  }
  cin >> s;
  cin >> m;
  n = (long long int)s.size();
  build(1, 0, n - 1);
  for (i = 0; i < m; i++) {
    cin >> x >> y;
    x--;
    y--;
    tri p1 = query(1, 0, n - 1, x, y);
    ans = 2 * p1.b;
    cout << ans << "\n";
  }
  return 0;
}
