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;
int tree[1000005], num[1000005], ac[1000005];
int query(int x) {
  int ans = 0;
  for (int i = x; i > 0; i -= i & -i) ans += tree[i];
  return ans;
}
void update(int x, int val) {
  for (int i = x; i < 1000005; i += i & -i) tree[i] += val;
}
int menor[1000005];
pair<int, int> S[1000005];
vector<pair<int, int> > C[1000005];
vector<int> O[1000005];
int main() {
  string s;
  while (cin >> s) {
    memset(num, -1, sizeof num);
    vector<int> v;
    int n = s.size();
    s = "." + s;
    for (int i = 1; i < s.size(); i++) {
      if (s[i] == '(')
        v.push_back(i);
      else if (v.size() > 0)
        num[i] = v.back(), v.pop_back();
    }
    for (int i = 1; i < s.size(); i++) ac[i] = ac[i - 1] + (num[i] > 0);
    int m;
    scanf("%d", &m);
    for (int i = 0; i < m; i++) {
      scanf("%d%d", &S[i].first, &S[i].second);
      O[S[i].first].push_back(i);
      C[S[i].second].push_back(make_pair(i, S[i].first));
    }
    int id, val;
    for (int i = n - 1; i >= 1; i--) {
      for (int j = 0; j < C[i].size(); j++) {
        id = C[i][j].first;
        val = C[i][j].second;
        menor[id] -= query(val - 1);
      }
      if (num[i] > 0) update(num[i], 1);
      for (int j = 0; j < O[i].size(); j++) {
        id = O[i][j];
        menor[id] += query(i - 1);
      }
    }
    int ans, l, r;
    for (int i = 0; i < m; i++) {
      l = S[i].first;
      r = S[i].second;
      ans = ac[r] - ac[l - 1] - menor[i];
      printf("%d\n", 2 * ans);
    }
  }
}
