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 long long ll;

int main()
{
    const int N=1000000;
    int n;
    scanf("%d",&n);
    char s[N+1];
    vector<array<int,2>> pos,neg;
    for(int o=1;o<=n;o++)
    {
        scanf("%s",s);
        int d=0;
        int m=0;
        for(int i=0;s[i]!='\0';i++)
        {
            if(s[i]=='(') d++;
            else d--;
            m=min(m,d);
        }
        if(d>=0) pos.push_back({-m,d});
        else neg.push_back({max(-m,-d),d});
    }
    sort(pos.begin(),pos.end());
    sort(neg.begin(),neg.end(),[](array<int,2> a,array<int,2> b){return (a[0]+a[1]>b[0]+b[1]);});
    int r=0;
    auto out=[]()->void
    {
        printf("No\n");
        exit(0);
    };
    for(auto [a,b]:pos)
    {
        if(r<a) out();
        r+=b;
    }
    for(auto [a,b]:neg)
    {
        if(r<a) out();
        r+=b;
    }
    if(r!=0) out();
    printf("Yes\n");
    return 0;
}
