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;
class info {
 public:
  int len, a, b;
  info operator+(info B) {
    info A;
    int t;
    t = min(this->a, B.b);
    A.len = this->len + B.len + t;
    A.a = this->a + B.a - t;
    A.b = this->b + B.b - t;
    return A;
  }
};
char bracket[1000006];
info tree[2100000];
void build(int pos, int L, int R) {
  if (L == R) {
    if (bracket[L - 1] == '(') {
      tree[pos].len = tree[pos].b = 0;
      tree[pos].a = 1;
    } else {
      tree[pos].len = tree[pos].a = 0;
      tree[pos].b = 1;
    }
    return;
  }
  int x = pos * 2, mid = (L + R) / 2;
  build(x, L, mid);
  build(x + 1, mid + 1, R);
  tree[pos] = tree[x] + tree[x + 1];
}
info query(int pos, int L, int R, int l, int r) {
  info A, B;
  if (L > r || R < l) {
    A.a = A.b = A.len = 0;
    return A;
  }
  if (L >= l && R <= r) return tree[pos];
  int x = pos * 2, mid = (L + R) / 2;
  A = query(x, L, mid, l, r);
  B = query(x + 1, mid + 1, R, l, r);
  return A + B;
}
int main() {
  int i, n, m, l, r;
  info A;
  scanf("%s", bracket);
  n = strlen(bracket);
  build(1, 1, n);
  scanf("%d", &m);
  for (i = 0; i < m; i++) {
    scanf("%d %d", &l, &r);
    A = query(1, 1, n, l, r);
    printf("%d\n", A.len * 2);
  }
  return 0;
}
