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 mod = 1E9 + 7;
const int maxn = 1E6 + 10;
const int inf = 0x3f3f3f3f;
string s;
int tot, now;
struct node {
  int a, b;
  int ans;
} tree[maxn << 2];
void build(int l, int r, int rt) {
  if (l == r) {
    if (s[l - 1] == '(')
      tree[rt].a = 1, tree[rt].b = 0;
    else
      tree[rt].b = 1, tree[rt].a = 0;
    tree[rt].ans = 0;
    return;
  }
  int mid = (l + r) >> 1;
  build(l, mid, rt << 1);
  build(mid + 1, r, rt << 1 | 1);
  int minn = min(tree[rt << 1].a, tree[rt << 1 | 1].b);
  tree[rt].a = tree[rt << 1].a + tree[rt << 1 | 1].a - minn;
  tree[rt].b = tree[rt << 1].b + tree[rt << 1 | 1].b - minn;
  tree[rt].ans = tree[rt << 1].ans + tree[rt << 1 | 1].ans + 2 * minn;
  return;
}
void query(int l, int r, int rt, int x, int y) {
  if (x <= l && r <= y) {
    tot += tree[rt].ans;
    int res = min(now, tree[rt].b);
    tot += res * 2;
    now = now + tree[rt].a - res;
    return;
  }
  int mid = (l + r) >> 1;
  if (mid >= x) query(l, mid, rt << 1, x, y);
  if (mid < y) query(mid + 1, r, rt << 1 | 1, x, y);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> s;
  build(1, (int)s.size(), 1);
  int num;
  cin >> num;
  while (num--) {
    int x, y;
    cin >> x >> y;
    tot = 0;
    now = 0;
    query(1, (int)s.size(), 1, x, y);
    cout << tot << endl;
  }
}
