Problem:
Score : 600 points

Problem Statement

A bracket sequence is a string that is one of the following:

An empty string;

The concatenation of (, A, and ) in this order, for some bracket sequence A ;

The concatenation of A and B in this order, for some non-empty bracket sequences A and B /

Given are N strings S_i. Can a bracket sequence be formed by concatenating all the N strings in some order?

Constraints

1 \leq N \leq 10^6

The total length of the strings S_i is at most 10^6.

S_i is a non-empty string consisting of ( and ).

Input

Input is given from Standard Input in the following format:

N
S_1
:
S_N

Output

If a bracket sequence can be formed by concatenating all the N strings in some order, print Yes; otherwise, print No.

Sample Input 1

2
)
(()

Sample Output 1

Yes

Concatenating (() and ) in this order forms a bracket sequence.

Sample Input 2

2
)(
()

Sample Output 2

No

Sample Input 3

4
((()))
((((((
))))))
()()()

Sample Output 3

Yes

Sample Input 4

3
(((
)
)

Sample Output 4

No

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

//////////////////////////////   Begin Macros

#define all(x) (x).begin(), (x).end()
#define rep(i, N) for (int i = 0; i < (int)(N); i++)
#define reprev(i, N) for (int i = (int)(N)-1; i >= 0; i--)
#define rep1(i, N) for (int i = 1; i <= (int)(N); i++)
#define rep1rev(i, N) for (int i = (int)(N); i >= 0; i--)
#define forbe(i, b, e) for (int i = (b); i < (e); i++)
#define forberev(i, b, e) for (int i = (e)-1; i >= (b); i--)
#define forfl(i, f, l) for (int i = (f); i <= (l); i++)
#define forflrev(i, f, l) for (int i = (l); i >= (f); i--)
using ll = long long;
using pii = pair<int, int>;
using pil = pair<int, ll>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;

template <typename T>
bool chmax(T &m, const T q)
{
  if (m < q)
  {
    m = q;
    return true;
  }
  else
    return false;
}
template <typename T>
bool chmin(T &m, const T q)
{
  if (m > q)
  {
    m = q;
    return true;
  }
  else
    return false;
}
template <typename T1, typename T2>
pair<T1, T2> operator+(const pair<T1, T2> &l, const pair<T1, T2> &r) { return {l.first + r.first, l.second + r.second}; }
template <typename T1, typename T2>
pair<T1, T2> operator-(const pair<T1, T2> &l, const pair<T1, T2> &r) { return {l.first - r.first, l.second - r.second}; }
template <typename T>
pair<T, T> operator*(const pair<T, T> &l, const T &r) { return {l.first * r, l.second * r}; }
template <typename T>
pair<T, T> operator/(const pair<T, T> &l, const T &r) { return {l.first / r, l.second / r}; }
template <typename T>
istream &operator>>(istream &is, vector<T> &vec)
{
  for (auto &v : vec)
    is >> v;
  return is;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &vec)
{
  os << "[";
  for (auto v : vec)
    os << v << ",";
  os << "]";
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const deque<T> &vec)
{
  os << "deq[";
  for (auto v : vec)
    os << v << ",";
  os << "]";
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &vec)
{
  os << "{";
  for (auto v : vec)
    os << v << ",";
  os << "}";
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const unordered_set<T> &vec)
{
  os << "{";
  for (auto v : vec)
    os << v << ",";
  os << "}";
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const multiset<T> &vec)
{
  os << "{";
  for (auto v : vec)
    os << v << ",";
  os << "}";
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const unordered_multiset<T> &vec)
{
  os << "{";
  for (auto v : vec)
    os << v << ",";
  os << "}";
  return os;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &pa)
{
  os << "(" << pa.first << "," << pa.second << ")";
  return os;
}
template <typename TK, typename TV>
ostream &operator<<(ostream &os, const map<TK, TV> &mp)
{
  os << "{";
  for (auto v : mp)
    os << v.first << "=>" << v.second << ",";
  os << "}";
  return os;
}
template <typename TK, typename TV>
ostream &operator<<(ostream &os, const unordered_map<TK, TV> &mp)
{
  os << "{";
  for (auto v : mp)
    os << v.first << "=>" << v.second << ",";
  os << "}";
  return os;
}

template <typename T>
void reset(vector<T> &v, const T reset_to)
{
  for (auto &x : v)
    x = reset_to;
}
inline int popcount(const unsigned int x) { return __builtin_popcount(x); }
#define dbg(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ") " << __FILE__ << endl;

ll nC2(ll n)
{
  return n * (n - 1) / 2;
}

const int intinf = numeric_limits<int>::max();
const ll llinf = numeric_limits<ll>::max();
const pii udlr[4] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};

//////////////////////////////   End Macros

void solve()
{
  int N;
  cin >> N;
  vector<string> S(N);
  rep(i, N) cin >> S[i];
  vector<ll> req(N), eff(N);
  ll s_total = 0;
  rep(i, N)
  {
    ll M = 0, m = 0;
    ll s = 0;
    for (char c : S[i])
    {
      if (c == '(')
      {
        s++;
        chmax(M, s);
      }
      else
      {
        s--;
        chmin(m, s);
      }
    }
    m *= -1;
    if (s >= 0)
      req[i] = m;
    else
      req[i] = M;
    eff[i] = s;
    s_total += s;
  }

  if (s_total != 0)
  {
    cout << "No" << endl;
    return;
  }

  vector<pll> req_eff_open, req_eff_close;
  rep(i, N) if (eff[i] >= 0)
  {
    req_eff_open.push_back({req[i], eff[i]});
    req_eff_close.push_back({req[i], eff[i]});
  }
  dbg(req_eff_open);
  dbg(req_eff_close);
  sort(all(req_eff_open));
  ll s_tmp_open = 0;
  for (auto [r, e] : req_eff_open)
  {
    if (s_tmp_open < r)
    {
      cout << "No" << endl;
      return;
    }
    s_tmp_open += e;
  }

  sort(all(req_eff_close));
  ll s_tmp_close = 0;
  for (auto [r, e] : req_eff_close)
  {
    if (s_tmp_close < r)
    {
      cout << "No" << endl;
      return;
    }
    s_tmp_close += e;
  }

  cout << "Yes" << endl;
}

int main()
{
  // srand(time(0));
  cout << fixed << setprecision(15);

  solve();
  return 0;
}