#include <bits/stdc++.h>
using namespace std;
int n;
struct node {
  int L;
  int R;
};
node tree[1000005 * 4];
int a[1000005];
string s;
void buildTree(int N, int B, int E) {
  if (B == E) {
    tree[N].L = 0;
    tree[N].R = 0;
    if (s[B] == '(') {
      tree[N].L = 1;
    } else {
      tree[N].R = 1;
    }
    return;
  }
  int M = (B + E) / 2;
  int L = 2 * N + 1;
  int R = L + 1;
  buildTree(L, B, M);
  buildTree(R, M + 1, E);
  tree[N].L = max(0, tree[L].L - tree[R].R) + tree[R].L;
  tree[N].R = max(0, tree[R].R - tree[L].L) + tree[L].R;
}
void update(int N, int B, int E, int i) {
  if (B > i || E < i) return;
  if (B == E) {
    swap(tree[N].L, tree[N].R);
    return;
  }
  int M = (B + E) / 2;
  int L = 2 * N + 1;
  int R = L + 1;
  update(L, B, M, i);
  update(R, M + 1, E, i);
  tree[N].L = max(0, tree[L].L - tree[R].R) + tree[R].L;
  tree[N].R = max(0, tree[R].R - tree[L].L) + tree[L].R;
}
int main() {
  cin >> n;
  cin >> s;
  buildTree(0, 0, n - 1);
  int ans = 0;
  for (int i = 0; i < n; i++) {
    update(0, 0, n - 1, i);
    ans += (tree[0].L + tree[0].R == 0);
    update(0, 0, n - 1, i);
  }
  cout << ans;
}

### Prompt

Create a solution in CPP for 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;
int n;
struct node {
  int L;
  int R;
};
node tree[1000005 * 4];
int a[1000005];
string s;
void buildTree(int N, int B, int E) {
  if (B == E) {
    tree[N].L = 0;
    tree[N].R = 0;
    if (s[B] == '(') {
      tree[N].L = 1;
    } else {
      tree[N].R = 1;
    }
    return;
  }
  int M = (B + E) / 2;
  int L = 2 * N + 1;
  int R = L + 1;
  buildTree(L, B, M);
  buildTree(R, M + 1, E);
  tree[N].L = max(0, tree[L].L - tree[R].R) + tree[R].L;
  tree[N].R = max(0, tree[R].R - tree[L].L) + tree[L].R;
}
void update(int N, int B, int E, int i) {
  if (B > i || E < i) return;
  if (B == E) {
    swap(tree[N].L, tree[N].R);
    return;
  }
  int M = (B + E) / 2;
  int L = 2 * N + 1;
  int R = L + 1;
  update(L, B, M, i);
  update(R, M + 1, E, i);
  tree[N].L = max(0, tree[L].L - tree[R].R) + tree[R].L;
  tree[N].R = max(0, tree[R].R - tree[L].L) + tree[L].R;
}
int main() {
  cin >> n;
  cin >> s;
  buildTree(0, 0, n - 1);
  int ans = 0;
  for (int i = 0; i < n; i++) {
    update(0, 0, n - 1, i);
    ans += (tree[0].L + tree[0].R == 0);
    update(0, 0, n - 1, i);
  }
  cout << ans;
}

```