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 ML = 1e6 + 5;
const int MM = 1e5 + 5;
int n, m;
char s[ML];
struct node {
  int opp, neg, res;
  node() { opp = neg = res = 0; }
  friend node operator+(const node &i, const node &j) {
    node now;
    int cur = min(i.opp, j.neg);
    now.res = i.res + j.res + cur;
    now.opp = i.opp + j.opp - cur;
    now.neg = i.neg + j.neg - cur;
    return now;
  }
} tree[ML << 2];
void build(int p, int pl, int pr) {
  if (pl == pr) {
    tree[p].res = 0;
    tree[p].opp = (s[pl] == '(');
    tree[p].neg = (s[pl] == ')');
    return;
  }
  int mid = (pl + pr) >> 1;
  build(p << 1, pl, mid);
  build(p << 1 | 1, mid + 1, pr);
  tree[p] = tree[p << 1] + tree[p << 1 | 1];
}
node ask(int p, int pl, int pr, int ql, int qr) {
  if (pl == ql && pr == qr) return tree[p];
  int mid = (pl + pr) >> 1;
  if (qr <= mid)
    return ask(p << 1, pl, mid, ql, qr);
  else if (ql > mid)
    return ask(p << 1 | 1, mid + 1, pr, ql, qr);
  else
    return ask(p << 1, pl, mid, ql, mid) +
           ask(p << 1 | 1, mid + 1, pr, mid + 1, qr);
}
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  build(1, 1, n);
  scanf("%d", &m);
  for (int i = 1; i <= m; ++i) {
    int ql, qr;
    scanf("%d%d", &ql, &qr);
    printf("%d\n", ask(1, 1, n, ql, qr).res * 2);
  }
  return 0;
}
