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;
char S[2000001];
int a[2000001], b[2000001], c[2000001];
void update(int ind, char C) {
  ind += N;
  if (C == '(')
    b[ind] = 1;
  else
    c[ind] = 1;
  for (ind /= 2; ind >= 1; ind /= 2) {
    int T = min(b[ind * 2], c[ind * 2 + 1]);
    a[ind] = a[ind * 2] + a[ind * 2 + 1] + T;
    b[ind] = b[ind * 2] + b[ind * 2 + 1] - T;
    c[ind] = c[ind * 2] + c[ind * 2 + 1] - T;
  }
}
int query(int L, int R) {
  L += N;
  R += N;
  vector<int> v1, v2;
  while (L <= R) {
    if (L % 2 == 1) {
      v1.push_back(L);
      L++;
    }
    if (R % 2 == 0) {
      v2.push_back(R);
      R--;
    }
    L /= 2;
    R /= 2;
  }
  int A = 0, B = 0, C = 0;
  for (int i = 0; i < v1.size(); i++) {
    int ind = v1[i];
    int T = min(B, c[ind]);
    A = A + a[ind] + T;
    B = B + b[ind] - T;
    C = C + c[ind] - T;
  }
  for (int i = v2.size() - 1; i >= 0; i--) {
    int ind = v2[i];
    int T = min(B, c[ind]);
    A = A + a[ind] + T;
    B = B + b[ind] - T;
    C = C + c[ind] - T;
  }
  return A;
}
int main() {
  scanf("%s", S);
  N = strlen(S);
  for (int i = 0; i < N; i++) update(i, S[i]);
  int Q;
  scanf("%d", &Q);
  while (Q--) {
    int L, R;
    scanf("%d%d", &L, &R);
    printf("%d\n", query(L - 1, R - 1) * 2);
  }
  return 0;
}
