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 node {
  int correct;
  int open;
  int close;
  node() { correct = open = close = 0; }
  node operator+(const node& rhs) const {
    node ans;
    ans.correct = correct + rhs.correct + min(open, rhs.close);
    ans.open = rhs.open + max(0, open - rhs.close);
    ans.close = close + max(0, rhs.close - open);
    return ans;
  }
  node operator=(const char c) {
    correct = 0;
    open = (c == '(');
    close = (c == ')');
    return *this;
  }
};
string s;
node seg[4004004];
node build(int id = 1, int l = 0, int r = s.size()) {
  if (l == r - 1) return seg[id] = s[l];
  int lChild = id * 2;
  int rChild = id * 2 + 1;
  int mid = (l + r) / 2;
  return seg[id] = build(lChild, l, mid) + build(rChild, mid, r);
}
node getAns(int x, int y, int id = 1, int l = 0, int r = s.size()) {
  if (x >= r || y <= l) return node();
  if (x <= l && y >= r) return seg[id];
  int lChild = id * 2;
  int rChild = id * 2 + 1;
  int mid = (l + r) / 2;
  return getAns(x, y, lChild, l, mid) + getAns(x, y, rChild, mid, r);
}
int main() {
  int m, l, r;
  cin >> s >> m;
  build();
  while (m--) {
    scanf("%d %d", &l, &r);
    printf("%d\n", getAns(l - 1, r).correct * 2);
  }
  return 0;
}
