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>
#pragma GCC optimize("O3")
using namespace std;
long long n;
char c[1000002];
struct aint {
  int lmax, st, dr;
};
aint v[4000002];
void build(int nod, int l, int r) {
  if (l == r) {
    v[nod] = {0, (c[l] == '('), (c[l] == ')')};
    return;
  }
  int mid = (l + r) / 2;
  build(nod * 2, l, mid);
  build(nod * 2 + 1, mid + 1, r);
  int t = min(v[nod * 2].st, v[nod * 2 + 1].dr);
  v[nod].lmax = v[nod * 2].lmax + v[nod * 2 + 1].lmax + 2 * t;
  v[nod].st = v[nod * 2].st + v[nod * 2 + 1].st - t;
  v[nod].dr = v[nod * 2].dr + v[nod * 2 + 1].dr - t;
}
int aa, bb;
void qu(int nod, int l, int r, int a, int b) {
  if (l > b || r < a) return;
  if (a <= l && r <= b) {
    int t = min(bb, v[nod].dr);
    aa = aa + v[nod].lmax + 2 * t;
    bb = bb + v[nod].st - t;
    return;
  }
  int mid = (l + r) / 2;
  qu(nod * 2, l, mid, a, b);
  qu(nod * 2 + 1, mid + 1, r, a, b);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> (c + 1);
  n = strlen(c + 1);
  build(1, 1, n);
  int q;
  cin >> q;
  for (int i = 1; i <= q; ++i) {
    int a, b;
    cin >> a >> b;
    aa = 0, bb = 0;
    qu(1, 1, n, a, b);
    cout << aa << '\n';
  }
  return 0;
}
