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 s[1001000];
int tree[1001000 << 2];
int lc[1001000 << 2];
int rc[1001000 << 2];
int n, m;
void init(int rt, int l, int r) {
  if (l == r) {
    if (s[l] == '(') {
      lc[rt] = 1;
    } else {
      rc[rt] = 1;
    }
    return;
  }
  int mid = (l + r) / 2;
  init(rt << 1, l, mid);
  init(rt << 1 | 1, mid + 1, r);
  tree[rt] =
      tree[rt << 1] + tree[rt << 1 | 1] +
      min(lc[rt << 1] - tree[rt << 1], rc[rt << 1 | 1] - tree[rt << 1 | 1]);
  lc[rt] = lc[rt << 1] + lc[rt << 1 | 1];
  rc[rt] = rc[rt << 1] + rc[rt << 1 | 1];
}
int query(int rt, int l, int r, int ql, int qr, int &lnum, int &rnum) {
  if (ql <= l && qr >= r) {
    lnum = lc[rt];
    rnum = rc[rt];
    return tree[rt];
  }
  int mid = (l + r) / 2;
  int t1 = 0, t2 = 0, t3 = 0, t4 = 0, t5 = 0, t6 = 0;
  if (ql <= mid) {
    t1 = query(rt << 1, l, mid, ql, qr, t3, t4);
  }
  if (qr > mid) {
    t2 = query(rt << 1 | 1, mid + 1, r, ql, qr, t5, t6);
  }
  lnum = t3 + t5;
  rnum = t4 + t6;
  return t1 + t2 + min(t3 - t1, t6 - t2);
}
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  init(1, 1, n);
  scanf("%d", &m);
  while (m--) {
    int ll, rr;
    int l, r;
    scanf("%d%d", &l, &r);
    printf("%d\n", 2 * query(1, 1, n, l, r, ll, rr));
  }
}
