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;
const int N = 1e6 + 10;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
int n;
char a[N];
int l[N],r[N],cnt,vis[N],f[N];
char q[N];
int main() {
  ios::sync_with_stdio(false);
  cin>>n;
  for(int i=1;i<=n;i++){
    f[i]=i;
    cin>>a+1;cnt=0;
    int n=strlen(a+1);
    for(int j=1;j<=n;j++){
      if(a[j]=='('){
        q[++cnt]='(';
      }else{
        if(q[cnt]=='(')--cnt;
        else q[++cnt]=')';
      }
    }
    for(int j=1;j<=cnt;j++){
      if(q[j]=='(')l[i]++;
      else r[i]++;
    }
 
  }
  int L=0,R=0;
  sort(f+1,f+1+n,[](int x,int y){
    if(l[x]>=r[x] && l[y]<r[y])return 1;
    if(l[x]<r[x] && l[y]>=r[y])return 0;
    if(l[x]>=r[x] && l[y]>=r[y])return (int)(r[x]<r[y]);
    if(l[x]<=r[x]&&l[y]<=r[y])return (int)(l[x]>l[y]);
  });
  for(int i=1;i<=n;i++){
    if(L<r[f[i]])return cout<<"No\n",0;
    L-=r[f[i]];
    L+=l[f[i]];
  }
  if(!L)cout<<"Yes\n";
  else cout<<"No\n";
  return 0;
}