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 <algorithm>
#include <bitset>
#include <cassert>
#include <cctype>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <functional>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <time.h>
#include <tuple>
#include <utility>
#include <vector>
#define ll long long
#define itn int
#define endl '\n'
#define co(ans) cout<<ans<<endl
#define COYE cout<<"YES"<<endl
#define COYe cout<<"Yes"<<endl
#define COye cout<<"yes"<<endl
#define CONO cout<<"NO"<<endl
#define CONo cout<<"No"<<endl
#define COno cout<<"no"<<endl
#define FORE(i,a) for(auto &i:a)
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define FFOR(i,a,b) for(int i=(a);i<=(b);++i)
#define REP(i,n)  FOR(i,0,n)
#define RREP(i,n)  FFOR(i,1,n)
#define PB push_back
#define MP make_pair
#define ALL(V) (V).begin(),(V).end()
#define SORT(V) sort((V).begin(),(V).end())
#define REVERSE(V) reverse((V).begin(),(V).end())
#define EACH(V,i) for(typeof((V).begin()) i=(V).begin();i!=(V).end();++i)
#define equals(a,b) (fabs((a)-(b))<EPS)
#define INF ((1LL<<62)-(1LL<<31))
#define EPS 1e-10
#define PI 3.141592653589793238
#define MOD 1000000007
#define MAX 5100000
using namespace std;
using Graph=vector<vector<int>>;
inline int toInt(string s){int v;istringstream sin(s);sin>>v;return v;}
template<class T>inline string toString(T x){ostringstream sout;sout<<x;return sout.str();}
template<class T>bool chmax(T &a,const T &b){if(a<b){a=b;return 1;}return 0;}
template<class T>bool chmin(T &a,const T &b){if(b<a){a=b;return 1;}return 0;}
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef long long LL;
const int dx[4]={0,1,0,-1},dy[4]={1,0,-1,0};
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}

int main(){
  ll N,ans=0;
  cin>>N;
  ll A[N],B[N];
  // A[i] S[i]の最小値
  // B[i] S[i]の値
  
  REP(i,N){
    A[i]=0;
    B[i]=0;
  }
  
  string S[N];
  REP(i,N) cin>>S[i];
  
  REP(i,N){
    ll cnt=0;
    REP(j,S[i].size()){
      if(S[i].at(j)=='(') cnt++;
      else cnt--;
      chmin(A[i],cnt);
    }
    B[i]=cnt;
  }
  
  vector<pair<int,int>> L,LL,LLL;
  bool CAN=true;
  
  REP(i,N){
    if(B[i]>=0ll){
      if(A[i]>=0ll) L.PB(MP(A[i],B[i]));
      else LL.PB(MP(A[i],B[i]));
    }
    else LLL.PB(MP(A[i],B[i]));
  }
  
  SORT(L);
  REVERSE(L);
  SORT(LL);
  REVERSE(LL);
  SORT(LLL);
  REVERSE(LLL);
  
  REP(i,L.size()){
    if(ans+L[i].first<0) CAN=false;
    ans+=L[i].second;
  }
  
  REP(i,LL.size()){
    if(ans+LL[i].first<0) CAN=false;
    ans+=LL[i].second;
  }
  
  REP(i,LLL.size()){
    if(ans+LLL[i].first<0) CAN=false;
    ans+=LLL[i].second;
  }
  
  if(ans!=0) CAN=false;
  if(CAN) cout<<"Yes"<<endl;
  else cout<<"No"<<endl;
  return 0;
}