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;
char t[1000005];
struct node {
  int O, C, sum;
};
node st[4 * 1000005];
void build(int n, int s, int e) {
  if (s == e) {
    if (t[s] == '(')
      st[n].O = 1;
    else
      st[n].C = 1;
  } else {
    int mid = (s + e) / 2;
    build(n + n, s, mid);
    build(n + n + 1, mid + 1, e);
    int tmp = min(st[n + n].O, st[n + n + 1].C);
    st[n].O = st[n + n].O + st[n + n + 1].O - tmp;
    st[n].C = st[n + n].C + st[n + n + 1].C - tmp;
    st[n].sum = st[n + n].sum + st[n + n + 1].sum + tmp;
  }
}
node query(int n, int s, int e, int l, int r) {
  if (s > r || l > e) return {0, 0, 0};
  if (l <= s && e <= r) return st[n];
  int mid = (s + e) / 2;
  node res = {0, 0, 0};
  node p1 = query(n + n, s, mid, l, r);
  node p2 = query(n + n + 1, mid + 1, e, l, r);
  int tmp = min(p1.O, p2.C);
  res.O = p1.O + p2.O - tmp;
  res.C = p1.C + p2.C - tmp;
  res.sum = p1.sum + p2.sum + tmp;
  return res;
}
int main() {
  scanf("%s", t + 1);
  int m = strlen(t + 1);
  build(1, 1, m);
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    printf("%d\n", 2 * query(1, 1, m, l, r).sum);
  }
}
