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 MOD = (int)1e9 + 7;
template <typename T>
static T aabs(T x) {
  if (x < 0) return -x;
  return x;
}
template <typename T>
static void read(T &S) {
  char K[1000009];
  scanf("%s", &K);
  S = K;
}
template <typename T, typename U>
static void amax(T &x, U y) {
  if (x < y) x = y;
}
template <typename T, typename U>
static void amin(T &x, U y) {
  if (x > y) x = y;
}
template <typename T, typename U>
static void aswap(T &x, U &y) {
  T temp;
  temp = x;
  x = y;
  y = temp;
}
template <typename T, typename U>
static long long gcd(T a, U b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
template <typename T, typename U>
static long long lcm(T a, U b) {
  return a / gcd(a, b) * b;
}
template <typename T, typename U>
static long long fpow(T n, U k, int p = MOD) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
struct Node {
  int optimal;
  int open;
  int close;
  Node() {}
  Node(int op, int o, int c) {
    optimal = op;
    open = o;
    close = c;
  }
};
const int N = 1000009;
Node st[N * 4];
string s;
Node operator+(const Node &left, const Node &right) {
  int temp;
  temp = min(left.open, right.close);
  Node ans;
  ans.optimal = left.optimal + right.optimal + temp;
  ans.open = left.open + right.open - temp;
  ans.close = left.close + right.close - temp;
  return ans;
}
void build(int id, int l, int r) {
  if (l == r) {
    if (s[l] == '(')
      st[id] = Node(0, 1, 0);
    else
      st[id] = Node(0, 0, 1);
    return;
  }
  int mid;
  mid = (l + r) / 2;
  build(id * 2, l, mid);
  build(id * 2 + 1, mid + 1, r);
  st[id] = st[id * 2] + st[id * 2 + 1];
}
Node query(int id, int l, int r, int u, int v) {
  if (r < u || l > v) return Node(0, 0, 0);
  if (l >= u && r <= v) return st[id];
  int mid;
  mid = (l + r) / 2;
  return query(id * 2, l, mid, u, v) + query(id * 2 + 1, mid + 1, r, u, v);
}
int main() {
  read(s);
  int n = s.size();
  s = " " + s + " ";
  build(1, 1, n);
  int q;
  scanf("%d", &q);
  while (q--) {
    int l, r;
    scanf("%d%d", &l, &r);
    Node ans;
    ans = query(1, 1, n, l, r);
    printf("%d\n", ans.optimal * 2);
  }
  return 0;
}
