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 MAXN = 1000010;
const int inf = 1 << 30;
int curSum[MAXN], n, m, a[3 * MAXN + 100];
string s;
void update(int node, int left, int right, int i, int v) {
  if (left >= right) {
    a[node] = v;
    return;
  }
  int mid = (left + right) / 2;
  if (i <= mid)
    update(node * 2, left, mid, i, v);
  else
    update(node * 2 + 1, mid + 1, right, i, v);
  a[node] = min(a[node * 2], a[2 * node + 1]);
}
int query(int node, int left, int right, int st, int fn) {
  if (st <= left && right <= fn) return a[node];
  int mid = (left + right) / 2, sol = inf;
  if (st <= mid) sol = min(sol, query(node * 2, left, mid, st, fn));
  if (fn > mid) sol = min(sol, query(node * 2 + 1, mid + 1, right, st, fn));
  return sol;
}
int abs(int a) {
  if (a < 0) return -a;
  return a;
}
int main() {
  cin >> s;
  n = s.size();
  for (int i = 0; i < n; ++i) {
    curSum[i + 1] = curSum[i];
    if (s[i] == '(')
      ++curSum[i + 1];
    else
      --curSum[i + 1];
    update(1, 1, n, i + 1, curSum[i + 1]);
  }
  cin >> m;
  for (int i = 1; i <= m; ++i) {
    int l, r;
    cin >> l >> r;
    int last = curSum[r] - curSum[l - 1];
    int minim = query(1, 1, n, l, r) - curSum[l - 1];
    int rez = 0;
    if (minim < 0) {
      minim = -minim;
      last += minim;
      rez += minim;
    }
    rez += abs(last);
    cout << (r - l + 1) - rez << "\n";
  }
  return 0;
}
