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;
struct st {
  int open, close, length;
} tree[4 * 1000100];
string str;
int Q, L, R, sz;
void makeTree(int node, int s, int e) {
  if (s == e) {
    if (str[s] == '(')
      tree[node].open = 1, tree[node].close = 0;
    else
      tree[node].open = 0, tree[node].close = 1;
    tree[node].length = 0;
    return;
  }
  int mid = (s + e) / 2;
  int left = node * 2;
  int right = node * 2 + 1;
  makeTree(left, s, mid);
  makeTree(right, mid + 1, e);
  int Pair = min(tree[left].open, tree[right].close);
  tree[node].length = tree[left].length + tree[right].length + (2 * Pair);
  tree[node].open = tree[left].open + tree[right].open - Pair;
  tree[node].close = tree[left].close + tree[right].close - Pair;
}
st query(int node, int s, int e, int x, int y) {
  st ret = {0, 0, 0};
  if (e < x || s > y) return ret;
  if (x <= s && e <= y) return tree[node];
  int mid = (s + e) / 2;
  int left = node * 2;
  int right = node * 2 + 1;
  st lf = query(left, s, mid, x, y);
  st rt = query(right, mid + 1, e, x, y);
  int Pair = min(lf.open, rt.close);
  ret.length = lf.length + rt.length + (2 * Pair);
  ret.open = lf.open + rt.open - Pair;
  ret.close = lf.close + rt.close - Pair;
  return ret;
}
int main() {
  cin >> str;
  cin >> Q;
  sz = str.size() - 1;
  makeTree(1, 0, sz);
  while (Q--) {
    cin >> L >> R;
    cout << query(1, 0, sz, L - 1, R - 1).length << endl;
  }
}
