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 n, m, M;
int lp[(int)1e6 + 7];
pair<int, int> ql[(int)1e5 + 7];
struct node {
  int a;
  int b;
  int c;
};
node D[(int)4e6 + 7];
node query(int a, int b, int w = 1, int p = 0, int k = M - 1) {
  int sr = (p + k) / 2;
  node ret;
  if (p == a && b == k)
    ret = D[w];
  else if (a <= sr && b <= sr)
    ret = query(a, b, w * 2, p, sr);
  else if (a >= sr + 1 && b >= sr + 1)
    ret = query(a, b, w * 2 + 1, sr + 1, k);
  else {
    node l = query(a, min(sr, b), w * 2, p, sr);
    node r = query(max(a, sr + 1), b, w * 2 + 1, sr + 1, k);
    int t = min(l.b, r.c);
    ret.a = l.a + r.a + t;
    ret.b = l.b + r.b - t;
    ret.c = l.c + r.c - t;
  }
  return ret;
}
int main() {
  string a;
  cin >> a;
  n = a.size();
  scanf("%d", &m);
  for (int i = 0; i < m; i++) scanf("%d %d", &ql[i].first, &ql[i].second);
  M = 1;
  while (M <= n) M *= 2;
  for (int i = M; i < M + n; i++) {
    D[i].a = 0;
    D[i].b = (a[i - M] == '(') ? 1 : 0;
    D[i].c = (a[i - M] == ')') ? 1 : 0;
  }
  for (int i = M - 1; i >= 1; i--) {
    int t = min(D[i * 2].b, D[i * 2 + 1].c);
    D[i].a = D[i * 2].a + D[i * 2 + 1].a + t;
    D[i].b = D[i * 2].b + D[i * 2 + 1].b - t;
    D[i].c = D[i * 2].c + D[i * 2 + 1].c - t;
  }
  for (int i = 0; i < m; i++)
    printf("%d\n", query(ql[i].first - 1, ql[i].second - 1).a * 2);
}
