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>
int B = 400;
using namespace std;
const long long inf = 1e17;
int MOD = 998244353;
const int mxn = 1000010;
int dx[] = {0, 1, 0, -1, 1, -1, 1, -1};
int dy[] = {1, 0, -1, 0, -1, 1, 1, -1};
long long n, m, k, ans;
string s;
struct node {
  int l, r, ans;
  node() { l = 0, r = 0; }
  node operator+(const node &oth) {
    node neu;
    neu.l = l + oth.l, neu.r = r + oth.r;
    int l_r = r - ans / 2, r_l = oth.l - oth.ans / 2;
    neu.ans += ans + oth.ans + 2 * max(0, min(l_r, r_l));
    return neu;
  }
} tree[4 * mxn];
void upd(int node, int l, int r, int i) {
  if (l == r) {
    int r_ = 0, l_ = 0, ans_ = 0;
    if (s[i - 1] == '(')
      r_++;
    else
      l_++;
    tree[node].l = l_, tree[node].r = r_, tree[node].ans = 0;
    return;
  }
  int mid = (l + r) / 2;
  if (i <= mid)
    upd(node * 2, l, mid, i);
  else
    upd(2 * node + 1, mid + 1, r, i);
  tree[node] = tree[2 * node] + tree[2 * node + 1];
}
int l_, r_, ans_ = 0;
struct node res;
void qry(int node, int l, int r, int ql, int qr, int ok = -1) {
  if (l > qr || r < ql || l > r) return;
  if (ql <= l && r <= qr) {
    res = res + tree[node];
    return;
  }
  int mid = (l + r) / 2;
  qry(2 * node, l, mid, ql, qr, 0);
  qry(2 * node + 1, mid + 1, r, ql, qr, 1);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  getline(cin, s);
  for (int i = 0; i < s.length(); i++) upd(1, 1, s.length(), i + 1);
  cin >> n;
  while (n--) {
    int x, y;
    cin >> x >> y;
    qry(1, 1, s.length(), x, y);
    cout << res.ans << endl;
    l_ = 0, r_ = 0, ans_ = 0;
    res.l = 0, res.r = 0, res.ans = 0;
  }
}
