A bracket sequence is a string that is one of the following:

1. An empty string;
2. The concatenation of `(`, A, and `)` in this order, for some bracket sequence A ;
3. 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`.

Examples

Input

2
)
(()


Output

Yes


Input

2
)(
()


Output

No


Input

4
((()))
((((((
))))))
()()()


Output

Yes


Input

3
(((
)
)


Output

No
#include <bits/stdc++.h>
using namespace std;
typedef pair<int,string> pis;

int n;
vector<pis> a,b;

int main()
{
    cin>>n;
    for(int i=0;i<n;i++)
    {
        string s,t;
        cin>>s;
        t=s;
        int cnt=0,mm=0x3f3f3f3f,mx=-0x3f3f3f3f;
        for(auto c:t)
        {
            if(c=='(') cnt++; else cnt--;
            mm=min(mm,cnt);
        }
        reverse(t.begin(),t.end()),cnt=0;
        for(auto c:t)
        {
            if(c=='(') cnt++; else cnt--;
            mx=max(mx,cnt);
        }
        if(cnt>=0) a.emplace_back(-mm,s); else b.emplace_back(-mx,s);
    }
    sort(a.begin(),a.end()),sort(b.begin(),b.end());
    int cnt=0,ok=1;
    for(auto s:a) for(auto c:s.second)
    {
        if(c=='(') cnt++; else cnt--;
        if(cnt<0) ok=0;
    }
    for(auto s:b) for(auto c:s.second)
    {
        if(c=='(') cnt++; else cnt--;
        if(cnt<0) ok=0;
    }
    if(ok&&cnt==0) printf("Yes"); else printf("No");
    return 0;
}
