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;
int n,a,b,A,B,aa,bb,aaa,bbb;
string s;
bool judge(int a,int b,int A,int B)
{
    A-=b;
    B-=a;
    return A>=0&&B>=0&&A==B;
}
vector<pair<int,int>>v;
bool cmp(pair<int,int>x,pair<int,int>y)
{
    return min(x.first,y.second)>min(x.second,y.first);
}
int main()
{
    srand(time(0));
    cin>>n;
    while(n--)
    {
        cin>>s;
        int l=0,r=0;
        for(int i=0;i<s.size();i++)
            if(s[i]=='(') r++;
        else
        {
            if(r) r--;
            else l++;
        }
        if(l&&r) v.push_back({l,r});
        else if(l) A+=l;
        else if(r) B+=r;
    }
    sort(v.begin(),v.end(),cmp);
    for(pair<int,int>x:v)
    {
        int l=x.first,r=x.second;
        int k=max(min(a,r),min(b,l));
        a=a+l-k;
        b=b+r-k;
    }
    if(judge(a,b,A,B))
        printf("Yes\n");
    else
    {
        printf("No\n");
    }
}
