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 FastIO {
 public:
  inline FastIO &operator<<(const int &a) {
    printf("%d", a);
    return *this;
  }
  inline FastIO &operator<<(const char *a) {
    printf("%s", a);
    return *this;
  }
  inline FastIO &operator>>(const int &a) {
    scanf("%d", &a);
    return *this;
  }
  inline FastIO &operator>>(const char *const &a) {
    scanf("%s", a);
    return *this;
  }
} hir;
struct node {
  int ans, op, cl;
  node() : ans(0), op(0), cl(0){};
};
const int N = 1e6 + 1, LG = 20;
node lr[N][LG];
char s[N];
node merg(node *a, node *b) {
  node ret;
  int mn = std::min(a->op, b->cl);
  ret.ans = 2 * mn + a->ans + b->ans;
  ret.op = a->op + b->op - mn;
  ret.cl = a->cl + b->cl - mn;
  return ret;
}
node get_par(int v, int cnt) {
  node ans;
  for (int p = 0; p < LG; p++)
    if (cnt & (1 << p)) {
      ans = merg(&lr[v][p], &ans);
      v = v - (1 << p);
    }
  return ans;
}
int main() {
  node a, b;
  hir >> s;
  int n = 0;
  while (s[n] != NULL) n++;
  for (int i = 0; i <= n; i++)
    for (int j = 0; j < LG; j++) lr[i][j] = node();
  for (int i = 0; i <= n; i++) {
    if (i == 0) continue;
    node cur;
    if (s[i - 1] == '(')
      cur.op++;
    else
      cur.cl++;
    lr[i][0] = cur;
    for (int bit = 1; bit < LG; bit++) {
      if (i - (1 << (bit - 1)) < 0) continue;
      lr[i][bit] = merg(&lr[i - (1 << (bit - 1))][bit - 1], &lr[i][bit - 1]);
    }
  }
  int q;
  hir >> q;
  for (int i = 0; i < q; i++) {
    int l, r;
    hir >> l >> r;
    l--;
    hir << get_par(r, r - l).ans << "\n";
  }
}
