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>
#define fi first
#define se second
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
#define srep(i,s,t) for (int i = s; i < t; ++i)
#define rng(a) a.begin(),a.end()
#define sz(x) (int)(x).size()
#define uni(x) x.erase(unique(rng(x)),x.end())
#define show(x) cout<<#x<<" = "<<x<<endl;
#define PQ(T) priority_queue<T,v(T),greater<T> >
#define newline puts("")
#define v(T) vector<T>
#define vv(T) v(v(T))
#define chmin(x,y) x=min(x,y)
#define chmax(x,y) x=max(x,y)
#define EPS (1e-10)
#define equals(a, b) (fabs((a)-(b)) < EPS)
using namespace std;
typedef long long int ll;
typedef pair<int,int> P;
typedef set<int> S;
typedef queue<int> Q;
typedef queue<P> QP;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<P> vp;
const ll LINF = 1001002003004005006ll;
const int INF = 1001001001;
const int MOD = 1000000007;
  
int main() {
  int N;
  cin >> N;
  v(string) ss(N);
  rep(i,N) {
    cin >> ss[i];
  }
  map<int,int> in,out;
  rep(i,N) {
    int dep = 0;
    int mi=0;
    rep(j,sz(ss[i])) {
      if(ss[i][j]=='(') {
        dep++;
      } else {
        dep--;
      }
      chmin(mi,dep);
    }
    int r=dep-mi;
    int l=-mi;
    if(l==r&&l==0) continue;
    if(l==r&&l>0) {
      in[l]++;
      continue;
    }
    in[l]++;
    out[r]++;
  }
  bool valid=true;
  for(auto k:in) {
    if(out[k.first]!=k.second) {
      valid=false;
      break;
    }
  }
  if(valid) {
    cout << "Yes" << endl;
  } else {
    cout << "No" << endl;
  }
  return 0;
}