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;
const int MAX = 1e6 + 9;
int tree[3 * MAX], op[3 * MAX], ans[3 * MAX], cl[3 * MAX];
string ar;
struct fin {
  int o, c, a;
};
void _merge(int node) {
  tree[node] = tree[2 * node] + tree[2 * node + 1];
  ans[node] =
      ans[2 * node] + ans[2 * node + 1] + min(op[2 * node], cl[2 * node + 1]);
  op[node] =
      op[2 * node] + op[2 * node + 1] - min(op[2 * node], cl[2 * node + 1]);
  cl[node] =
      cl[2 * node] + cl[2 * node + 1] - min(op[2 * node], cl[2 * node + 1]);
}
void build(int node, int st, int en) {
  if (st == en) {
    if (ar[st - 1] == '(')
      tree[node] = 1, op[node] = 1;
    else
      tree[node] = -1, cl[node] = 1;
    return;
  }
  int mid = (st + en) / 2;
  build(2 * node, st, mid);
  build(2 * node + 1, mid + 1, en);
  _merge(node);
}
fin query(int node, int st, int en, int l, int r) {
  if (st >= l && en <= r) {
    fin temp;
    temp.a = ans[node];
    temp.c = cl[node];
    temp.o = op[node];
    return temp;
  }
  if (l > en || r < st) {
    fin temp;
    temp.a = 0;
    temp.o = 0;
    temp.c = 0;
    return temp;
  }
  int mid = (st + en) / 2;
  fin f1 = query(2 * node, st, mid, l, r);
  fin f2 = query(2 * node + 1, mid + 1, en, l, r);
  fin f;
  f.a = (f1.a + f2.a) + min(f1.o, f2.c);
  f.o = (f1.o + f2.o) - min(f1.o, f2.c);
  f.c = (f1.c + f2.c) - min(f1.o, f2.c);
  return f;
}
int main(void) {
  memset(tree, 0, sizeof(tree));
  memset(ans, 0, sizeof(ans));
  memset(op, 0, sizeof(op));
  memset(cl, 0, sizeof(cl));
  cin >> ar;
  int n = ar.size();
  build(1, 1, n);
  int m;
  scanf("%d", &m);
  while (m--) {
    int in1, in2;
    scanf("%d %d", &in1, &in2);
    printf("%d\n", query(1, 1, n, in1, in2).a * 2);
  }
  return 0;
}
