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;
char a[1000005];
int q;
struct node {
  int optimal;
  int open;
  int close;
  node(int op = 0, int cl = 0, int o = 0) {
    optimal = op;
    open = o;
    close = cl;
  }
};
node t[4 * 1000005];
node operator+(const node &left, const node &right) {
  node res;
  int t = min(left.open, right.close);
  res.optimal = left.optimal + right.optimal + t;
  res.close = left.close + right.close - t;
  res.open = left.open + right.open - t;
  return res;
}
void update(int k, int l, int r) {
  if (l == r) {
    if (a[l] == '(')
      t[k] = node(0, 1, 0);
    else
      t[k] = node(0, 0, 1);
    return;
  }
  int m = (l + r) / 2;
  update(k * 2, l, m);
  update(k * 2 + 1, m + 1, r);
  t[k] = t[k * 2 + 1] + t[k * 2];
}
node query(int k, int l, int r, int u, int v) {
  if (v < l || u > r) return node(0, 0, 0);
  if (u <= l && v >= r) {
    return t[k];
  }
  int m = (l + r) / 2;
  return (query(k * 2 + 1, m + 1, r, u, v) + query(k * 2, l, m, u, v));
}
int main() {
  scanf("%s", a + 1);
  n = strlen(a + 1);
  update(1, 1, n);
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    printf("%d\n", query(1, 1, n, u, v).optimal * 2);
  }
}
