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;
inline long long msb(long long val) { return 63 - __builtin_clzll(val); }
inline int msb(int val) { return 31 - __builtin_clz(val); }
template <class T>
bool chmin(T& a, const T& b) {
  return a > b ? a = b, true : false;
}
template <class T>
bool chmax(T& a, const T& b) {
  return a < b ? a = b, true : false;
}
const long long mod = 1e9 + 7, N = 1e6 + 5;
long long a[N], n, m, k;
struct node {
  long long ans, l, r;
  node() { ans = l = r = 0; }
} t[N * 2];
inline node merge(const node& a, const node& b) {
  node res;
  long long t = min(a.l, b.r);
  res.ans = max({a.ans + b.ans + t * 2});
  res.l = a.l + b.l - t;
  res.r = a.r + b.r - t;
  return res;
}
long long get(long long l, long long r) {
  node resl, resr;
  for (l += n, r += n; l <= r; l >>= 1, r >>= 1) {
    if (l & 1) resl = merge(resl, t[l++]);
    if (!(r & 1)) resr = merge(t[r--], resr);
  }
  return merge(resl, resr).ans;
}
void solve(long long test_case) {
  long long i, j, Q;
  string s;
  cin >> s;
  n = s.size();
  for (i = 0; i < n; i++) {
    if (s[i] == '(') t[i + n].l = 1;
    if (s[i] == ')') t[i + n].r = 1;
  }
  for (i = n - 1; i > 0; i--) t[i] = merge(t[i << 1], t[i << 1 | 1]);
  cin >> Q;
  while (Q--) {
    long long l, r;
    cin >> l >> r;
    l--, r--;
    cout << get(l, r) << '\n';
  }
  return;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  solve(1);
  return 0;
}
