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)

#pragma GCC target ("avx")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#define _USE_MATH_DEFINES
#include<iostream>
#include<string>
#include<queue>
#include<cmath>
#include<map>
#include<set>
#include<list>
#include<iomanip>
#include<vector>
#include<random>
#include<functional>
#include<algorithm>
#include<stack>
#include<cstdio>
#include<cstring>
#include<bitset>
#include<unordered_map>
#include<climits>
#include<fstream>
#include<complex>
#include<time.h>
#include<cassert>
#include<functional>
#include<numeric>
#include<tuple>
using namespace std;
using ll = long long;
using ld = long double;
#define int long long
#define all(a) (a).begin(),(a).end()
#define fs first
#define sc second
#define xx first
#define yy second.first
#define zz second.second
#define H pair<int, int>
#define P pair<int, pair<int, int>>
#define Q(i,j,k) mkp(i,mkp(j,k))
#define rng(i,s,n) for(int i = (s) ; i < (n) ; i++)
#define rep(i,n) rng(i, 0, (n))
#define mkp make_pair
#define vec vector
#define pb emplace_back
#define crdcomp(b) sort(all((b)));(b).erase(unique(all((b))),(b).end())
#define getidx(b,i) lower_bound(all(b),(i))-b.begin()
#define ssp(i,n) (i==(int)(n)-1?"\n":" ")
#define ctoi(c) (int)(c-'0')
#define itoc(c) (char)(c+'0')
#define pp(x,y) pb(H{x,y})
#define ppp(x,y,z) pb(Q(x,y,z))
#define cyes printf("Yes\n")
#define cno printf("No\n")
#define cdf(n) int quetimes_=(n);rep(qq123_,quetimes_)
#define gcj printf("Case #%lld: ",qq123_+1)
#define readv(a,n) a.resize(n,0);rep(i,(n)) a[i]=read()
//#define endl "\n"
constexpr int mod = 1e9 + 7;
constexpr int Mod = 998244353;
constexpr int MXN = 500000 + 100;
constexpr ld EPS = 1e-10;
constexpr ll inf = 3 * 1e18;
constexpr int Inf = 15 * 1e8;
const vec<int>dx{ -1,1,0,0 }, dy{ 0,0,-1,1 };
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; }
ll read() { ll u, k = scanf("%lld", &u); return u; }
string reads() { string s; cin >> s; return s; }
H readh(bool g = 0) { H u; int k = scanf("%lld %lld", &u.fs, &u.sc); if (g) u.fs--, u.sc--; return u; }
bool inarea(H t, int h, int w) { return 0 <= t.fs && t.fs < h && 0 <= t.sc && t.sc < w; }
ll gcd(ll i, ll j) { return j ? gcd(j, i % j) : i; }
ll mod_pow(ll x, ll n, ll p = mod) {
    ll res = 1; x %= p;
    while (n > 0) {
        if (n & 1) res = res * x % p;
        x = x * x % p;
        n >>= 1;
    }
    return res;
}//x^n%p
ll bitcount(ll x) {
    int sum = 0; for (int i = 0; i < 60; i++)if ((1ll << i) & x) sum++;
    return sum;
}
constexpr int fn_ = 5000005;
ll fact_[fn_], comp_[fn_];
ll comb(ll x, ll y, ll m) {
    if (!fact_[0]) {
        fact_[0] = 1; comp_[0] = 1;
        for (int i = 1; i < fn_; i++)
            fact_[i] = fact_[i - 1] * i % m;
        comp_[fn_ - 1] = mod_pow(fact_[fn_ - 1], m - 2, m);
        for (int i = fn_ - 2; i > 0; i--)
            comp_[i] = comp_[i + 1] * (i + 1) % m;
    }
    if (x < y) return 0;
    return fact_[x] * comp_[x - y] % m * comp_[y] % m;
}
//--------------------------------------------------------------
signed main() {
    int n = read();
    vec<pair<string, int>>v;
    rep(i, n) {
        string s; cin >> s;
        vec<int>b(s.size(), 0);
        if (s[0] == '(') b[0] = 1;
        else b[0] = -1;
        rng(i, 1, s.size()) {
            if (s[i] == '(') b[i] = b[i - 1] + 1;
            else b[i] = b[i - 1] - 1;
        }
        int rrr = b[0];
        rep(i, s.size()) chmin(rrr, b[i]);
        v.pb(s, rrr);
    }
    sort(all(v), [](auto a, auto b) {
        if (a.sc != b.sc) return a.sc > b.sc;;
        return a.fs < b.fs;
    });
    string h = "";
    rep(i, v.size()) h += v[i].fs;
    int sum = 0;
    rep(i, h.size()) {
        if (h[i] == '(') sum++;
        else sum--;
        if (sum < 0) {
            cno;
            return 0;
        }
    }
    if (!sum) cyes;
    else cno;
}