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;
string s;
struct point {
  int a, b, c;
};
point seg[2100005];
void make_tree(int cur, int st, int en) {
  if (st == en) {
    seg[cur].a = 0;
    if (s[st] == '(') {
      seg[cur].b = 1;
      seg[cur].c = 0;
    } else {
      seg[cur].c = 1;
      seg[cur].b = 0;
    }
    return;
  }
  make_tree(2 * cur, st, (st + en) / 2);
  make_tree(2 * cur + 1, (st + en) / 2 + 1, en);
  int k = min(seg[2 * cur].b, seg[2 * cur + 1].c);
  seg[cur].a = seg[2 * cur].a + seg[2 * cur + 1].a + 2 * k;
  seg[cur].b = seg[2 * cur].b + seg[2 * cur + 1].b - k;
  seg[cur].c = seg[2 * cur].c + seg[2 * cur + 1].c - k;
}
point query(int cur, int st, int en, int x, int y) {
  if (x > en || y < st) {
    point u;
    u.a = 0, u.b = 0, u.c = 0;
    return u;
  }
  if (st >= x && en <= y) {
    return seg[cur];
  }
  point le = query(2 * cur, st, (st + en) / 2, x, y);
  point ri = query(2 * cur + 1, (st + en) / 2 + 1, en, x, y);
  int j = min(le.b, ri.c);
  point uo;
  uo.a = le.a + ri.a + 2 * j;
  uo.b = le.b + ri.b - j;
  uo.c = le.c + ri.c - j;
  return uo;
}
int main() {
  cin >> s;
  int n = s.length();
  make_tree(1, 0, n - 1);
  int q;
  cin >> q;
  while (q--) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    point l = query(1, 0, n - 1, x, y);
    cout << l.a << "\n";
  }
  return 0;
}
