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<iostream>
#include<algorithm>
#include<vector>
#include<string>
#include<set>
#include<queue>
#include<stack>
#include<bitset>
#include<unordered_map>
#include<functional>
#include<map>
#include<iomanip>
#include<limits>
#include<unordered_set>
#include<cmath>
#include <numeric>
#include <array>
#include<utility>
#include <complex>
#define M_PI 3.141592653589793238
using namespace std;
long long p9 = 998244353;
long long p1 = 1000000007;
#define upperbound(v,val) upper_bound(v.begin(),v.end(),val)-v.begin()
#define lowerbound(v,val) lower_bound(v.begin(),v.end(),val)-v.begin()
#define int long long
#define ll long long
#define vel vector<ll>
#define vvel vector<vel>
#define rep(i,n) for(int i=0;i<n;i++)
#define sor(v) sort(v.begin(),v.end())
#define mmax(a,b) a=max(a,b)
#define mmin(a,b) a=min(a,b)
#define mkp(a,b) make_pair(a,b)
#define pin pair<ll,ll>
#define qin pair<pin,int>
#define V vector
#define Endl endl
#define veb vector<bool>
#define fcout cout << fixed << setprecision(15)
#define rev(s) reverse(s.begin(),s.end())
#define lower(h,val) (lower_bound(h.begin(),h.end(),val)-h.begin())
#define upper(h,val) (upper_bound(h.begin(),h.end(),val)-h.begin())
#define vveb V<veb>
#define omajinai cin.tie(0);ios::sync_with_stdio(false);
#define endl "\n"
#define pb push_back
vel kai;
vel inv_kai;
vel inv;
int root(int x, vel& pa) {
	if (pa[x] == -1) { return x; }
	int ans = root(pa[x], pa); pa[x] = ans;
	return ans;
}
bool mar(int x, int y, vel& pa) {
	x = root(x, pa);
	y = root(y, pa);
	if (x != y) { pa[x] = y; }
	return (x != y);
}
int gcd(int x, int y) {
	if (x < y) { return gcd(y, x); }
	if (y == 0) { return x; }
	return gcd(y, x % y);
}
int lcm(ll x, ll y) {
	x = abs(x); y = abs(y);
	return x * (y / gcd(x, y));
}
long long modinv(long long a, long long m) {
	long long b = m, u = 1, v = 0;
	while (b) {
		long long t = a / b;
		a -= t * b; swap(a, b);
		u -= t * v; swap(u, v);
	}
	u %= m;
	if (u < 0) u += m;
	return u;
}
void make_inv(int max_inv, int p) {
	inv = vel(max_inv + 1, 1);
	for (int i = 2; i <= max_inv; i++) {
		inv[i] = p - ((p / i) * inv[p % i]) % p;
	}
}
void make_kai(int max_kai, int p) {
	kai = vel(max_kai + 1, 1);
	inv_kai = kai;
	make_inv(max_kai, p);
	rep(i, max_kai) {
		kai[i + 1] = kai[i] * (i + 1); kai[i + 1] %= p;
		inv_kai[i + 1] = inv_kai[i] * inv[i + 1]; inv_kai[i + 1] %= p;
	}
}
int com(int n, int r, int p) {
	if ((n < 0) || (r < 0) || (r > n)) { return 0; }
	int ans = (kai[n] * inv_kai[r]) % p;
	return (ans * inv_kai[n - r]) % p;
}
int per(int n, int r, int p) {
	if ((n < 0) || (r < 0) || (r > n)) { return 0; }
	return (kai[n] * inv_kai[n - r]) % p;
}
vel uni(vel x) {
	if (x.size() == 0) { return x; }
	sor(x);
	int n = x.size();
	vel ans(1, x[0]);
	for (int j = 1; j < n; j++) {
		if (x[j - 1] != x[j]) { ans.push_back(x[j]); }
	}
	x = ans;
	return x;
}
vel dijk(V<V<pin>>& way, int st, int inf) {
	int n = way.size();
	vel dist(n, inf); dist[st] = 0;
	priority_queue<pin, vector<pin>, greater<pin>> pq;
	pq.push(mkp(0, st));
	veb is_checked(n, false);
	while (!pq.empty()) {
		pin x = pq.top(); pq.pop();
		int pot = x.second;
		if (!is_checked[pot]) {
			is_checked[pot] = true;
			for (auto y : way[pot]) {
				int nex_dist = x.first + y.second;
				int nex_pot = y.first;
				if (dist[nex_pot] > nex_dist) {
					dist[nex_pot] = nex_dist;
					pq.push(mkp(nex_dist, y.first));
				}
			}
		}
	}
	return dist;
}
void make_tree(vvel& chi, vel& par, int n) {
	V<V<pin>> way(n);
	rep(i, n - 1) {
		int a, b; cin >> a >> b; a--; b--;
		way[a].push_back(mkp(b, 1));
		way[b].push_back(mkp(a, 1));
	}
	vel dist = dijk(way, 0, n + 1);
	par = vel(n, -1);
	chi = vvel(n);
	rep(i, n) {
		for (auto nex : way[i]) {
			int pot = nex.first;
			if (dist[pot] > dist[i]) { chi[i].push_back(pot); }
			else { par[i] = pot; }
		}
	}
}
void pri(vel& v) {
	if (v.size() == 0) { return; }
	cout << v[0];
	rep(i, v.size() - 1) { cout << " " << v[i + 1]; }
	cout << endl;
	return;
}
int modpow(int a, int n, int p) {
	if (n == 0) { return 1; }
	int m = n / 2;
	int x = modpow(a, n / 2, p);
	x *= x; x %= p;
	if (n % 2 == 1) { x *= a; x %= p; }
	return x;
}
vel dx = { 0,0,-1,1 };
vel dy = { 1,-1,0,0 };
#define ui long long
int h, w;
int ind(int a,int b) { return a * w + b; }
vvel disj_min(vel& v) {
	int n = v.size();
	vvel ret(22, vel(n));
	ret[0] = v;
	rep(i, 21) {
		rep(j, n) {
			int nex = j + (1 << i);
			if (nex < n) {
				ret[i + 1][j] = min(ret[i][j], ret[i][nex]);
			}
			else {
				ret[i + 1][j] = ret[i][j];
			}
		}
	}
	return ret;
}
int find_min(vvel& dv, int l, int r) {
	int i = 21;
	while (l + (1 << i) > r) {
		i--;
	}
	while (i >= 0) {
		if (dv[i][l] > dv[i][r - (1 << i)]) {
			l = r - (1 << i);
		}
		else {
			r = l + (1 << i);
		}
		i--;
	}
	return l;
}
void sol(V<string> &t, int i, string u) {
	t.push_back(u);
	if (i < u.size()) {
		sol(t, i + 1, u);
		u[i] = '.';
		sol(t, i + 1, u);
	}
}
V<V<pin>> dbl(V<pin>& v) {
	V<V<pin>> ans(20, V<pin>(v));
	int n = v.size();
	rep(i, 19) {
		rep(j, n) {
			ans[i + 1][j].first = ans[i][ans[i][j].first].first;
			ans[i + 1][j].second = max(ans[i][j].second, ans[i][ans[i][j].first].second);
		}
	}
	return ans;
}
int lca(int s, int t, int diff,V<V<pin>>& pa) {
	if (diff < 0) { return lca(t, s, -diff, pa); }
	int ans = 0;
	rep(i, 19) {
		if ((diff & (1 << i)) != 0) {
			mmax(ans, pa[i][s].second);
			s = pa[i][s].first;
		}
	}
	for (int i = 19; i >= 0; i--) {
		if (pa[i][s] != pa[i][t]) {
			mmax(ans, pa[i][s].second);
			s = pa[i][s].first;
			mmax(ans,pa[i][t].second);
			t = pa[i][t].first;
		}
	}
	if (s != t) {
		mmax(ans, pa[0][s].second);
		mmax(ans, pa[0][t].second);
	}
	return ans;
}
signed main() {
	omajinai;
	int n; cin >> n;
	V<pin> pl, mi;
	int sum = 0;
	rep(i, n) {
		string s; cin >> s;
		int d = 0; int mn = 0;
		int sz = s.size();
		rep(j, sz) {
			if (s[j] == '(') { d++; }
			else { d--; }
			mmin(mn, d);
		}
		if (d >= 0) { pl.push_back(mkp(mn, d)); }
		else { mi.push_back(mkp(mn-d, -d)); }
		sum += d;
	}
	if (sum != 0) { cout << "No" << endl; return 0; }
	sor(pl); sor(mi);
	int qs = 0;
	for (auto x : pl) {
		if (qs < -x.first) { cout << "No" << endl; return 0; }
		qs += x.second;
	}
	qs = 0;
	for (auto x : mi) {
		if (qs < -x.first) { cout << "No" << endl; return 0; }
		qs += x.second;
	}
	cout << "Yes" << endl;
	return 0;
}