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 str[1000005];
vector<pair<int, int>> qr[1000005];
int fw[1000005], lnk[1000005], n, res[100005];
void fw_up(int first, int second, int v) {
  first = max(1, first);
  second = min(second, n);
  if (first > second) return;
  while (first <= n) {
    fw[first] += v;
    first += first & -first;
  }
  second++;
  while (second <= n) {
    fw[second] -= v;
    second += second & -second;
  }
}
int fw_get(int first) {
  int sum = 0;
  if (first < 1 || first > n) return 0;
  while (first > 0) {
    sum += fw[first];
    first -= first & -first;
  }
  return sum;
}
int main() {
  int t, i, j, k, m, first, second, idx;
  vector<int> stk;
  scanf("%s %d", str + 1, &t);
  for (i = 1; i <= t; i++) {
    scanf("%d %d", &first, &second);
    qr[first].emplace_back(second, i);
  }
  n = strlen(str + 1);
  for (i = 1; i <= n; i++) {
    fw_up(i, i, fw_get(i - 1));
    if (str[i] == '(') {
      stk.emplace_back(i);
    } else {
      if (!stk.empty()) {
        lnk[stk.back()] = i;
        lnk[i] = stk.back();
        stk.pop_back();
        fw_up(i, i, 1);
      }
    }
  }
  for (i = 1; i <= n; i++) {
    while (!qr[i].empty()) {
      first = qr[i].back().first;
      idx = qr[i].back().second;
      qr[i].pop_back();
      res[idx] = fw_get(first);
    }
    if (str[i] == '(') {
      if (lnk[i]) {
        fw_up(lnk[i], n, -1);
      }
    }
  }
  for (i = 1; i <= t; i++) {
    printf("%d\n", res[i] * 2);
  }
  return 0;
}
