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 N = 1002000;
int n;
char a[N];
struct data {
  int x, sr, sl;
};
struct node {
  int l, r;
  data t;
  node *c[2];
} mem[N * 2], *cur = mem, *root;
void update(data &x, const data &l, const data &r) {
  static int tmp;
  tmp = min(l.sl, r.sr);
  x.x = l.x + r.x + tmp;
  x.sl = l.sl + r.sl - tmp;
  x.sr = l.sr + r.sr - tmp;
}
node *init(int l, int r) {
  node *p = cur++;
  p->c[0] = p->c[1] = NULL;
  p->l = l;
  p->r = r;
  if (l == r) {
    p->t.x = 0;
    p->t.sl = a[l] == '(';
    p->t.sr = a[l] == ')';
    return p;
  }
  int mid = l + r >> 1;
  p->c[0] = init(l, mid);
  p->c[1] = init(mid + 1, r);
  update(p->t, p->c[0]->t, p->c[1]->t);
  return p;
}
data get(node *x, int l, int r) {
  if (l <= x->l && r >= x->r) {
    return x->t;
  }
  int mid = x->l + x->r >> 1;
  if (r <= mid) return get(x->c[0], l, r);
  if (l > mid) return get(x->c[1], l, r);
  data t, lt, rt;
  lt = get(x->c[0], l, r);
  rt = get(x->c[1], l, r);
  update(t, lt, rt);
  return t;
}
int main() {
  int i, m, l, r;
  scanf("%s", a + 1);
  n = strlen(a + 1);
  scanf("%d", &m);
  root = init(1, n);
  while (m--) {
    scanf("%d%d", &l, &r);
    printf("%d\n", get(root, l, r).x * 2);
  }
}
