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, m, i, j, kcnt, kl, kr, x, y;
char s[1000005];
struct f {
  int l, r, cnt;
} tree[4000005];
void build(int l, int r, int ind) {
  if (l == r) {
    if (s[l - 1] == '(')
      tree[ind].l++;
    else
      tree[ind].r++;
    return;
  }
  int mid = (l + r) / 2;
  build(l, mid, ind * 2);
  build(mid + 1, r, ind * 2 + 1);
  int val = min(tree[ind * 2].l, tree[ind * 2 + 1].r);
  tree[ind].cnt = tree[ind * 2].cnt + tree[ind * 2 + 1].cnt + val;
  tree[ind].l = tree[ind * 2].l + tree[ind * 2 + 1].l - val;
  tree[ind].r = tree[ind * 2].r + tree[ind * 2 + 1].r - val;
}
void query(int l, int r, int ind) {
  if (x <= l && r <= y) {
    int val = min(kl, tree[ind].r);
    kcnt += tree[ind].cnt + val;
    kl += tree[ind].l - val;
    kr += tree[ind].r - val;
    return;
  }
  int mid = (l + r) / 2;
  if (x <= mid) query(l, mid, ind * 2);
  if (y > mid) query(mid + 1, r, ind * 2 + 1);
}
int main() {
  scanf("%s", s);
  n = strlen(s);
  build(1, n, 1);
  scanf("%d", &m);
  for (i = 0; i < m; i++) {
    scanf("%d%d", &x, &y);
    kcnt = 0;
    kl = 0;
    kr = 0;
    query(1, n, 1);
    printf("%d\n", kcnt * 2);
  }
}
