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)

/**
 *    author:  otera    
**/
#include<iostream>
#include<string> 
#include<cstdio>
#include<cstring>
#include<vector>
#include<cmath>
#include<algorithm> 
#include<functional>
#include<iomanip>
#include<queue>
#include<deque>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<complex>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
#include<cassert>
using namespace std;

//#define int long long
typedef long long ll;
typedef unsigned long long ul;
typedef unsigned int ui;
typedef long double ld;
const int inf=1e9+7;
const ll INF=1LL<<60 ;
const ll mod=1e9+7 ;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
typedef complex<ld> Point;
const ld eps = 1e-8;
const ld pi = acos(-1.0);
typedef pair<int, int> P;
typedef pair<ld, ld> LDP;
typedef pair<ll, ll> LP;
#define fr first
#define sc second
#define all(c) c.begin(),c.end()
#define pb push_back
#define debug(x)  cerr << #x << " = " << (x) << endl;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

void solve() {
	int n; cin >> n;
    vector<string> s(n);
    rep(i, n) {
        cin >> s[i];
    }
    vector<int> mi(n, 0), fi(n, 0);
    vector<P> fo, bc;
    rep(i, n) {
        int sz = (int)s[i].size();
        int now = 0;
        rep(j, sz) {
            if(s[i][j] == '(') ++ now;
            else -- now;

            chmin(mi[i], now);
        }
        fi[i] = now;
        if(fi[i] >= 0) {
            fo.pb({mi[i], fi[i]});
        } else {
            bc.pb({mi[i], fi[i]});
        }
    }
    sort(all(fo));
    sort(all(bc));
    reverse(all(fo));
    int cur = 0;
    rep(i, (int)fo.size()) {
        if(cur + fo[i].fr < 0) {
            cout << "No" << endl;
            return;
        }
        cur += fo[i].sc;
    }
    rep(i, (int)bc.size()) {
        if(cur + bc[i].fr < 0) {
            cout << "No" << endl;
            return;
        }
        cur += bc[i].sc;
    }
    cout << "Yes" << endl;
}

signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	//cout << fixed << setprecision(10);
	//int t; cin >> t; rep(i, t)solve();
	solve();
    return 0;
}