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;
inline int minx(int a, int b) { return a < b ? a : b; }
struct data {
  int rf, lf, val;
  data() {}
  data(int rf, int lf, int val) : lf(lf), rf(rf), val(val) {}
  data(char c) {
    val = 0;
    if (c == '(') {
      rf = 1;
      lf = 0;
    } else {
      rf = 0;
      lf = 1;
    }
  }
};
data merge(const data& d1, const data& d2) {
  int mn = minx(d1.rf, d2.lf);
  return {d1.rf + d2.rf - mn, d1.lf + d2.lf - mn, d1.val + d2.val + 2 * mn};
}
class SegmentTree {
 public:
  int n;
  vector<data> t;
  SegmentTree(const string& s) {
    int sz = s.length();
    t.resize(4 * sz);
    build(s, 0, 0, sz - 1);
  }
  void build(const string& s, int v, int tl, int tr) {
    if (tl == tr) {
      t[v] = data(s[tl]);
    } else {
      int tm = (tr + tl) >> 1;
      build(s, 2 * v + 1, tl, tm);
      build(s, 2 * v + 2, tm + 1, tr);
      t[v] = merge(t[2 * v + 1], t[2 * v + 2]);
    }
  }
  data query(int v, int tl, int tr, int l, int r) {
    if (l == tl && r == tr) return t[v];
    int tm = (tr + tl) >> 1;
    if (l <= tm) {
      if (r <= tm)
        return query(2 * v + 1, tl, tm, l, r);
      else
        return merge(query(2 * v + 1, tl, tm, l, tm),
                     query(2 * v + 2, tm + 1, tr, tm + 1, r));
    } else {
      return query(2 * v + 2, tm + 1, tr, l, r);
    }
  }
};
int main() {
  string s;
  int m, l, r;
  cin >> s >> m;
  int sz = s.length();
  SegmentTree st(s);
  for (int i = 0; i < m; i++) {
    cin >> l >> r;
    --l;
    --r;
    cout << st.query(0, 0, sz - 1, l, r).val << '\n';
  }
  return 0;
}
