#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
int n, pmn[N], mn[N];
char s[N];
void dbg() { puts(""); }
template <typename T, typename... R>
void dbg(const T &f, const R &...r) {
  cout << f << " ";
  dbg(r...);
}
template <typename Type>
inline void read(Type &xx) {
  Type f = 1;
  char ch;
  xx = 0;
  for (ch = getchar(); ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') f = -1;
  for (; ch >= '0' && ch <= '9'; ch = getchar()) xx = xx * 10 + ch - '0';
  xx *= f;
}
int solve() {
  for (int i = (0); i < (n); ++i)
    mn[i] = (i ? mn[i - 1] : 0) + (s[i] == '(' ? 1 : -1);
  int sum = mn[n - 1];
  mn[n] = n + 1;
  for (int i = (n - 1); i >= (0); --i) mn[i] = min(mn[i], mn[i + 1]);
  for (int i = (0); i < (n); ++i)
    pmn[i] = (i ? pmn[i - 1] : 0) + (s[i] == '(' ? 1 : -1);
  for (int i = (1); i < (n); ++i) pmn[i] = min(pmn[i], pmn[i - 1]);
  int ans = 0;
  if (sum == 2) {
    for (int i = (0); i < (n); ++i)
      if (s[i] == '(' && mn[i] >= 2 && (!i || pmn[i - 1] >= 0)) ++ans;
  }
  if (sum == -2) {
    for (int i = (0); i < (n); ++i)
      if (s[i] == ')' && mn[i] >= -2 && (!i || pmn[i - 1] >= 0)) ++ans;
  }
  return ans;
}
int main(int argc, char **argv) {
  read(n);
  scanf("%s", s);
  printf("%d\n", solve());
  return 0;
}

### Prompt

Please create a solution in Cpp to the following problem:

You are given a bracket sequence s consisting of n opening '(' and closing ')' brackets.

A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.

You can change the type of some bracket s_i. It means that if s_i =  ')' then you can change it to '(' and vice versa.

Your task is to calculate the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.

Input

The first line of the input contains one integer n (1 ≤ n ≤ 10^6) — the length of the bracket sequence.

The second line of the input contains the string s consisting of n opening '(' and closing ')' brackets.

Output

Print one integer — the number of positions i such that if you change the type of the i-th bracket, then the resulting bracket sequence becomes regular.

Examples

Input


6
(((())


Output


3


Input


6
()()()


Output


0


Input


1
)


Output


0


Input


8
)))(((((


Output


0

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
int n, pmn[N], mn[N];
char s[N];
void dbg() { puts(""); }
template <typename T, typename... R>
void dbg(const T &f, const R &...r) {
  cout << f << " ";
  dbg(r...);
}
template <typename Type>
inline void read(Type &xx) {
  Type f = 1;
  char ch;
  xx = 0;
  for (ch = getchar(); ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') f = -1;
  for (; ch >= '0' && ch <= '9'; ch = getchar()) xx = xx * 10 + ch - '0';
  xx *= f;
}
int solve() {
  for (int i = (0); i < (n); ++i)
    mn[i] = (i ? mn[i - 1] : 0) + (s[i] == '(' ? 1 : -1);
  int sum = mn[n - 1];
  mn[n] = n + 1;
  for (int i = (n - 1); i >= (0); --i) mn[i] = min(mn[i], mn[i + 1]);
  for (int i = (0); i < (n); ++i)
    pmn[i] = (i ? pmn[i - 1] : 0) + (s[i] == '(' ? 1 : -1);
  for (int i = (1); i < (n); ++i) pmn[i] = min(pmn[i], pmn[i - 1]);
  int ans = 0;
  if (sum == 2) {
    for (int i = (0); i < (n); ++i)
      if (s[i] == '(' && mn[i] >= 2 && (!i || pmn[i - 1] >= 0)) ++ans;
  }
  if (sum == -2) {
    for (int i = (0); i < (n); ++i)
      if (s[i] == ')' && mn[i] >= -2 && (!i || pmn[i - 1] >= 0)) ++ans;
  }
  return ans;
}
int main(int argc, char **argv) {
  read(n);
  scanf("%s", s);
  printf("%d\n", solve());
  return 0;
}

```