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;
struct Info {
  int seqLen;
  int cnt[2];
};
const int N = 1000000;
int n;
int a[N + 5];
Info sgt[4 * N + 5];
Info operator+(Info info1, Info info2) {
  Info sumInfo;
  int t = min(info1.cnt[0], info2.cnt[1]);
  sumInfo.seqLen = info1.seqLen + info2.seqLen + t * 2;
  sumInfo.cnt[0] = info1.cnt[0] + info2.cnt[0] - t;
  sumInfo.cnt[1] = info1.cnt[1] + info2.cnt[1] - t;
  return sumInfo;
}
void updateSgtTree(int i, int l, int r, int pos) {
  if (r < i || l > i) return;
  if (l == r) {
    sgt[pos].seqLen = 0;
    sgt[pos].cnt[a[i]] = 1;
    return;
  }
  int mid = (l + r) / 2;
  updateSgtTree(i, l, mid, pos * 2);
  updateSgtTree(i, mid + 1, r, pos * 2 + 1);
  Info info1 = sgt[pos * 2];
  Info info2 = sgt[pos * 2 + 1];
  sgt[pos] = info1 + info2;
}
Info querySgtTree(int ql, int qr, int l, int r, int pos) {
  if (r < ql || l > qr) {
    Info empty;
    empty.seqLen = 0;
    empty.cnt[0] = 0;
    empty.cnt[1] = 0;
    return empty;
  }
  if (l >= ql && r <= qr) return sgt[pos];
  int mid = (l + r) / 2;
  return querySgtTree(ql, qr, l, mid, pos * 2) +
         querySgtTree(ql, qr, mid + 1, r, pos * 2 + 1);
}
int main() {
  string s;
  cin >> s;
  n = s.length();
  for (int i = 0; i < n; i++)
    if (s[i] == '(')
      a[i + 1] = 0;
    else
      a[i + 1] = 1;
  for (int i = 1; i <= n; i++) updateSgtTree(i, 1, n, 1);
  int m;
  scanf("%d", &m);
  while (m--) {
    int ql, qr;
    scanf("%d %d", &ql, &qr);
    Info t = querySgtTree(ql, qr, 1, n, 1);
    printf("%d\n", t.seqLen);
  }
  return 0;
}
