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;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
const long long N = 1e6;
long long t[4 * N][3];
long long n;
string s;
void build(long long v, long long l, long long r) {
  if (l == r) {
    t[v][0] = t[v][1] = t[v][2] = 0;
    t[v][s[l] - '(']++;
  } else {
    long long m = (l + r) >> 1;
    build(v << 1, l, m);
    build((v << 1) | 1, m + 1, r);
    long long mn = min(t[v << 1][0] - t[v << 1][2],
                       t[(v << 1) ^ 1][1] - t[(v << 1) ^ 1][2]);
    t[v][2] = t[(v << 1)][2] + t[(v << 1) ^ 1][2] + mn;
    t[v][0] = t[(v << 1)][0] + t[(v << 1) ^ 1][0];
    t[v][1] = t[v << 1][1] + t[(v << 1) ^ 1][1];
  }
}
vector<long long> get(long long v, long long tl, long long tr, long long l,
                      long long r) {
  if (l > tr || tl > r) return {0, 0, 0};
  if (l <= tl && tr <= r) return {t[v][0], t[v][1], t[v][2]};
  long long m = (tl + tr) / 2;
  auto x = get(v << 1, tl, m, l, r);
  auto y = get((v << 1) ^ 1, m + 1, tr, l, r);
  vector<long long> res(3, 0);
  long long mn = min(x[0] - x[2], y[1] - y[2]);
  res[2] = x[2] + y[2] + mn;
  res[0] = x[0] + y[0];
  res[1] = x[1] + y[1];
  return res;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  ;
  cin >> s;
  n = s.size();
  build(1, 0, n - 1);
  long long m;
  cin >> m;
  while (m--) {
    long long l, r;
    cin >> l >> r;
    cout << get(1, 0, n - 1, l - 1, r - 1)[2] * 2 << endl;
  }
}
