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: Accepted)

#include <bits/stdc++.h>
#include <unistd.h>
#include <sys/time.h>
#include <stdlib.h>
 
using namespace std;
 
typedef long long ll;
typedef pair<ll,ll> P;
typedef pair<P, ll> T;
 
long long int INF = 2e18;
const ll fact_table = 1200008;
double Pi = 3.1415926535897932384626;

priority_queue <ll> pql;
priority_queue <P> pqp;
priority_queue <P> bag;
//big priority queue
priority_queue <ll,vector<ll>,greater<ll> > pqls;
priority_queue <P,vector<P>,greater<P> > pqps;
//small priority queue
//top pop
 
ll dx[8]={1,0,-1,0,1,1,-1,-1};
ll dy[8]={0,1,0,-1,1,-1,-1,1};
char dir[] = "DRUL";
//↓,→,↑,←

#define endl "\n"
#ifdef ENJAPMA
    #undef endl
#endif
#define p(x) cout<<x<<endl;
#define el cout<<endl;
#define pe(x) cout<<x<<" ";
#define ps(x) cout<<fixed<<setprecision(25)<<x<<endl;
#define pu(x) cout<<(x);
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define CLEAR(a) a = decltype(a)(); 
#define pc(x) cout << x << ",";
 
const ll mod = 998244353ll;
// const ll mod = 1000000007ll;
  
ll mypow(ll number1,ll number2, ll mod){
    if(number2 == 0){
        return 1ll;
    }else{
        ll number3 = mypow(number1,number2 / 2, mod);
        number3 *= number3;
        number3 %= mod;
        if(number2%2==1){
            number3 *= number1;
            number3 %= mod;
        }
        return number3;
    }
}
void YES(bool condition){
    if(condition){
        p("YES");
    }else{
        p("NO");
    }
    return;
}
void Yes(bool condition){
    if(condition){
        p("Yes");
    }else{
        p("No");
    }
    return;
}


ll fact[fact_table + 5],rfact[fact_table + 5]; 
 
void c3_init(){
    fact[0] = rfact[0] = 1;
    for(ll i=1; i<=fact_table; i++){
        fact[i] = (fact[i-1]*i) % mod;
    }
    rfact[fact_table] = mypow(fact[fact_table],mod - 2, mod);
    for(ll i=fact_table; i>=1; i--){
       rfact[i-1] = rfact[i] * i;
       rfact[i-1] %= mod;
    }
    return;}
ll c3(ll n,ll r){
    assert(n >= r);
    assert(n >= 0);
    assert(r >= 0);
    return (((fact[n] * rfact[r]) % mod ) * rfact[n-r]) % mod;}

bool multicase = false;

ll n,m,num,a,b,c,d,e,q;

ll x[200005], y[100005], z[100005];
ll k, ans;
string s[1000005];

vector<P> up, down, even;

void solve(){
    bool dame = false;
    cin >> n;
    ll cnt = 0, sum = 0;
    for(ll i=0;i<n;i++){
        ll mindepth = 0;
        cin >> s[i];
        cnt = 0;
        ll m = s[i].size();
        for(ll j=0;j<m;j++){
            if(s[i][j] == '('){
                cnt ++;
                mindepth = min(cnt, mindepth);
            }
            if(s[i][j] == ')'){
                cnt --;
                mindepth = min(cnt, mindepth);
            }
        }
        sum += cnt;
        if(cnt > 0){
            up.pb(P(-mindepth, cnt));
        }else if(cnt < 0){
            down.pb(P(-mindepth + cnt, cnt));
        }else{
            even.pb(P(-mindepth, cnt));
        }
    }

    if(sum != 0)dame = true;

    sort(up.begin(), up.end());
    sort(down.begin(), down.end());

    ll height = 0;
    for(ll i=0;i<up.size();i++){
        if(dame)break;
        if(height < up[i].first)dame = true;
        height += up[i].second;
    }
    for(ll i=0;i<even.size();i++){
        if(dame)break;
        if(height < even[i].first)dame = true;
    }

    height = 0;

    for(ll i=0;i<down.size();i++){
        if(dame)break;
        if(height < down[i].first)dame = true;
        height -= down[i].second;
    }
    Yes(!dame);
    return;
}

int main(){
    // init();
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    ll q, testcase = 1;
    if(multicase){
        cin >> q;
    }else{
        q = 1; 
    }
    while(q--){
        // pu("Case ");pu("#");pu(testcase);pu(": ");
        solve();
        testcase++;
    }
    // solve();
    return 0;
}