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 node {
  int val, o, c;
  void join(node a, node b) {
    int tmp = min(a.o, b.c);
    val = a.val + b.val + tmp;
    o = a.o + b.o - tmp;
    c = a.c + b.c - tmp;
  }
};
node T[4 * 1000000];
char A[1000010];
void init(int b, int e, int nodo) {
  if (b == e) {
    T[nodo].val = 0;
    T[nodo].o = 0;
    T[nodo].c = 0;
    if (A[b] == '(')
      T[nodo].o++;
    else
      T[nodo].c++;
    return;
  }
  int L = 2 * nodo + 1, R = 2 * nodo + 2, mid = (b + e) / 2;
  init(b, mid, L);
  init(mid + 1, e, R);
  T[nodo].join(T[L], T[R]);
}
node query(int b, int e, int nodo, int i, int j) {
  if (i <= b && e <= j) {
    return T[nodo];
  }
  int L = 2 * nodo + 1, R = 2 * nodo + 2, mid = (b + e) / 2;
  if (j <= mid)
    return query(b, mid, L, i, j);
  else if (i > mid)
    return query(mid + 1, e, R, i, j);
  else {
    node aa, bb, cc;
    aa = query(b, mid, L, i, j);
    bb = query(mid + 1, e, R, i, j);
    cc.join(aa, bb);
    return cc;
  }
}
int main() {
  scanf("%s", &A);
  int n = strlen(A);
  init(0, n - 1, 0);
  int q;
  cin >> q;
  while (q--) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    cout << 2 * query(0, n - 1, 0, a, b).val << endl;
  }
  return 0;
}
