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 fx[] = {+1, -1, +0, +0}, fy[] = {+0, +0, +1, -1};
const long long mod = 1e9 + 7, INF = 1e18, mx = 1e6;
int Case = 1, cas = 0;
string S;
struct exo {
  int Open, Close, SubLen;
  exo() {}
  exo(int a, int b, int c) { Open = a, Close = b, SubLen = c; }
} tree[3 * mx + 5];
exo Combine(exo p, exo q) {
  int mn = min(p.Open, q.Close);
  int a = p.Open + q.Open - mn;
  int b = p.Close + q.Close - mn;
  int c = p.SubLen + q.SubLen + 2 * mn;
  return exo(a, b, c);
}
void build(int node, int L, int R) {
  if (L > R) return;
  if (L == R) {
    if (S[L] == '(')
      tree[node] = exo(1, 0, 0);
    else
      tree[node] = exo(0, 1, 0);
    return;
  }
  int mid = (L + R) >> 1;
  int Left = node << 1;
  int Right = Left | 1;
  build(Left, L, mid);
  build(Right, mid + 1, R);
  tree[node] = Combine(tree[Left], tree[Right]);
}
exo query(int node, int L, int R, int i, int j) {
  if (i > R || j < L) return exo(0, 0, 0);
  if (i <= L && R <= j) return tree[node];
  int mid = (L + R) >> 1;
  int Left = node << 1;
  int Right = Left | 1;
  exo p = query(Left, L, mid, i, j);
  exo q = query(Right, mid + 1, R, i, j);
  return Combine(p, q);
}
int main() {
  ios_base ::sync_with_stdio(0), cin.tie(0);
  ;
  cin >> S;
  int n = S.size();
  build(1, 0, n - 1);
  int Q;
  cin >> Q;
  while (Q--) {
    int L, R;
    cin >> L >> R;
    L--, R--;
    cout << query(1, 0, n - 1, L, R).SubLen << endl;
  }
  return 0;
}
