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;
const int MX = 1000005, INF = 1000000000, mod = 1000000007;
const long long int LINF = 1000000000000000000ll;
const double eps = 1e-10;
const int dx[] = {-1, 0, 1, 0}, dy[] = {0, -1, 0, 1};
char in[MX];
string s;
int l[MX], r[MX], c[MX];
vector<int> h[MX];
int main() {
  int n, m;
  scanf("%s%d", in, &m);
  s = in;
  n = s.size();
  stack<int> t;
  for (int i = 0; i < n; i++) {
    if (s[i] == '(') {
      t.push(i);
    } else {
      if (!t.empty()) {
        l[t.top() + 1] = c[t.top() + 1] = 1;
        r[i + 1] = c[i + 1] = 1;
        t.pop();
      }
    }
  }
  for (int i = 0; i < n; i++) {
    l[i + 1] += l[i];
    r[i + 1] += r[i];
    c[i + 1] += c[i];
  }
  int mx = 0;
  for (int i = 1; i <= n; i++) {
    if (l[i] - l[i - 1] == 1) {
      h[l[i] - r[i]].push_back(i);
      mx = max(mx, l[i] - r[i]);
    }
  }
  for (int i = 0; i < mx + 1; i++) h[i].push_back(INF);
  int a, b;
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &a, &b);
    a--;
    int x = c[b] - c[a];
    int y = l[b] - r[a];
    int ar1 = l[a + 1] - r[a + 1];
    if (l[a + 1] - l[a] == 1) ar1--;
    int ar2 = l[b] - r[b];
    int al = 0, ar = min(ar1, ar2) + 1, ac, ax, ay;
    while (al + 1 < ar) {
      ac = (al + ar) / 2;
      ax = lower_bound(h[ac].begin(), h[ac].end(), a + 1) - h[ac].begin();
      ay = lower_bound(h[ac].begin(), h[ac].end(), b + 1) - h[ac].begin();
      if (ay == ax)
        al = ac;
      else
        ar = ac;
    }
    y -= al;
    int d = y * 2 - x;
    printf("%d\n", x - d);
  }
  return 0;
}
