Problem: Petya's friends made him a birthday present — a bracket sequence. Petya was quite disappointed with his gift, because he dreamed of correct bracket sequence, yet he told his friends nothing about his dreams and decided to fix present himself. 

To make everything right, Petya is going to move at most one bracket from its original place in the sequence to any other position. Reversing the bracket (e.g. turning "(" into ")" or vice versa) isn't allowed. 

We remind that bracket sequence s is called correct if: 

  * s is empty; 
  * s is equal to "(t)", where t is correct bracket sequence; 
  * s is equal to t_1 t_2, i.e. concatenation of t_1 and t_2, where t_1 and t_2 are correct bracket sequences. 



For example, "(()())", "()" are correct, while ")(" and "())" are not. Help Petya to fix his birthday present and understand whether he can move one bracket so that the sequence becomes correct.

Input

First of line of input contains a single number n (1 ≤ n ≤ 200 000) — length of the sequence which Petya received for his birthday.

Second line of the input contains bracket sequence of length n, containing symbols "(" and ")".

Output

Print "Yes" if Petya can make his sequence correct moving at most one bracket. Otherwise print "No".

Examples

Input


2
)(


Output


Yes


Input


3
(()


Output


No


Input


2
()


Output


Yes


Input


10
)))))(((((


Output


No

Note

In the first example, Petya can move first bracket to the end, thus turning the sequence into "()", which is correct bracket sequence.

In the second example, there is no way to move at most one bracket so that the sequence becomes correct.

In the third example, the sequence is already correct and there's no need to move brackets.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long mod = pow(10, 9) + 7;
long long n, m, a[1000005], i, j, t, k;
long long prime[1000005] = {0};
vector<long long> vprime;
long long max3(long long a, long long b, long long c) {
  return max(a, max(b, c));
}
long long min3(long long a, long long b, long long c) {
  return min(a, min(b, c));
}
long long inv(long long x1, long long y1) {
  if (y1 == 0) return 1;
  long long temp = inv(x1, y1 / 2);
  temp = (temp * temp) % mod;
  if (y1 % 2) temp = (temp * x1) % mod;
  return temp;
}
long long inv1(long long a1, long long m1) {
  long long m0 = m1, x0 = 0, y0 = 1;
  if (m1 == 1) return 0;
  while (a1 > 1) {
    long long q = a1 / m1;
    long long temp = m1;
    m1 = a1 % m1;
    a1 = temp;
    temp = y0;
    y0 = x0 - q * y0;
    x0 = temp;
  }
  x0 = (x0 + m0) % m0;
  return x0;
}
void sieve() {
  for (i = 2; i < 1000; i++) {
    if (prime[i] == 0) {
      for (j = i * i; j <= 1000000; j += i) prime[j] = 1;
    }
  }
  for (i = 2; i < 1000001; i++) {
    if (prime[i] == 0) vprime.push_back(i);
  }
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL);
  cin >> n;
  string s;
  cin >> s;
  if (n % 2 == 1) {
    cout << "No";
    return 0;
  }
  long long a1 = 0, b1 = 0;
  for (i = 0; i < n; i++) {
    if (s[i] == '(')
      a1++;
    else
      b1++;
  }
  if (a1 != b1) {
    cout << "No";
    return 0;
  }
  long long c = 0, ne = 0;
  for (i = 0; i < n; i++) {
    if (ne > 1) {
      cout << "No";
      return 0;
    }
    if (s[i] == '(')
      c++;
    else {
      if (c == 0)
        ne++;
      else
        c--;
    }
  }
  if (ne > 1)
    cout << "No";
  else
    cout << "Yes";
  return 0;
}