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>
const int inf = 100100;
const int maxn = (int)1e6 + 10;
const int mod = (int)1e9 + 7;
using namespace std;
struct node {
  int pref, suff, sum;
} t[4 * maxn];
string second;
int n;
node merge(node a, node b) {
  node c;
  int x = min(a.suff, b.pref);
  c.sum = a.sum + b.sum + 2 * x;
  c.pref = a.pref + b.pref - x;
  c.suff = a.suff - x + b.suff;
  return c;
}
node get(int v, int tl, int tr, int l, int r) {
  if (tl > r || tr < l) return {0, 0, 0};
  if (l <= tl && tr <= r) return t[v];
  int mid = (tl + tr) >> 1;
  return merge(get(v * 2, tl, mid, l, r), get(v * 2 + 1, mid + 1, tr, l, r));
}
void build(int v, int l, int r) {
  if (l == r) {
    if (second[l] == '(')
      t[v] = {0, 1, 0};
    else
      t[v] = {1, 0, 0};
  } else {
    int mid = (l + r) >> 1;
    build(v * 2, l, mid);
    build(v * 2 + 1, mid + 1, r);
    t[v] = merge(t[v * 2], t[v * 2 + 1]);
  }
}
void solve() {
  cin >> second;
  n = second.size();
  build(1, 0, n - 1);
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    int l, r;
    cin >> l >> r;
    cout << get(1, 0, n - 1, l - 1, r - 1).sum << "\n";
  }
}
int main() {
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}
