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>
namespace ringo {
template <class T>
inline void read(T &x) {
  x = 0;
  register char c = getchar();
  register bool f = 0;
  while (!isdigit(c)) f ^= c == '-', c = getchar();
  while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
  if (f) x = -x;
}
template <class T>
inline void print(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) print(x / 10);
  putchar('0' + x % 10);
}
template <class T>
inline void print(T x, char c) {
  print(x), putchar(c);
}
const int N = 1e6 + 10;
int n, m;
char a[N];
struct info {
  int L, S, R;
  inline info() {}
  inline info(int l, int s, int r) { L = l, R = r, S = s; }
  inline info(char c) { c == '(' ? (L = 1, R = S = 0) : (R = 1, L = S = 0); }
  friend inline info operator+(const info &a, const info &b) {
    int min = std::min(a.L, b.R);
    return info(a.L + b.L - min, a.S + b.S + min, a.R + b.R - min);
  }
};
struct node {
  int l, r, mid;
  info x;
} p[N << 2];
void build(int u, int l, int r) {
  p[u].l = l, p[u].r = r, p[u].mid = (l + r) >> 1;
  if (l == r)
    p[u].x = info(a[l]);
  else {
    build(u << 1, l, p[u].mid);
    build(u << 1 | 1, p[u].mid + 1, r);
    p[u].x = p[u << 1].x + p[u << 1 | 1].x;
  }
}
info query(int u, int l, int r) {
  if (l == p[u].l && r == p[u].r) return p[u].x;
  if (r <= p[u].mid) return query(u << 1, l, r);
  if (l > p[u].mid) return query(u << 1 | 1, l, r);
  return query(u << 1, l, p[u].mid) + query(u << 1 | 1, p[u].mid + 1, r);
}
void main() {
  scanf("%s", a + 1), n = strlen(a + 1), read(m);
  build(1, 1, n);
  for (int i = 1, l, r; i <= m; i++) {
    read(l), read(r);
    print(query(1, l, r).S << 1, '\n');
  }
}
}  // namespace ringo
signed main() { return ringo::main(), 0; }
