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 a, b, c;
  Node() {}
  Node(int _a, int _b, int _c) {
    a = _a;
    b = _b;
    c = _c;
  }
};
Node tree[4000000];
int N, Q;
char ins[1000003];
void ini(int x, int l, int r) {
  if (l == r) {
    if (ins[l] == '(') {
      tree[x] = {1, 0, 0};
    } else {
      tree[x] = {0, 1, 0};
    }
    return;
  }
  int m = (l + r) >> 1;
  ini((x << 1), l, m);
  ini((x << 1) + 1, m + 1, r);
  Node ln = tree[(x << 1)], rn = tree[(x << 1) + 1];
  int t = min(ln.a, rn.b);
  tree[x] = {ln.a + rn.a - t, ln.b + rn.b - t, ln.c + rn.c + t};
}
Node get(int x, int l, int r, int L, int R) {
  if (r < L || R < l) return {0, 0, 0};
  if (L <= l && r <= R) return tree[x];
  int m = (l + r) >> 1;
  Node ln = get((x << 1), l, m, L, R), rn = get((x << 1) + 1, m + 1, r, L, R);
  int t = min(ln.a, rn.b);
  return Node(ln.a + rn.a - t, ln.b + rn.b - t, ln.c + rn.c + t);
}
int main() {
  scanf(" %s", ins + 1);
  N = strlen(ins + 1) + 1;
  ini(1, 1, N);
  scanf("%d", &Q);
  while (Q--) {
    int a, b;
    scanf("%d %d", &a, &b);
    Node t = get(1, 1, N, a, b);
    printf("%d\n", 2 * t.c);
  }
}
