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

import java.io.*;
import java.util.*;
public class Main {
	public static void main(String[] args) throws Exception{
		// TODO Auto-generated method stub
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		String[] cmd=br.readLine().split(" ");
		int n=Integer.valueOf(cmd[0]);
		int[][] arr=new int[n][2];
		for(int i=0;i<n;i++)
		{
			StringBuilder s=new StringBuilder(br.readLine());
			Stack<Character> st=new Stack<>();
			int cnt=0;
			for(int j=0;j<s.length();j++)
			{
				if(s.charAt(j)=='(')
					st.push('(');
				else
				{
					if(st.empty())
						cnt++;
					else
						st.pop();
				}
			}
			arr[i][1]=st.size();
			arr[i][0]=cnt;
		}
		StringBuilder f=new StringBuilder("");
		for(int i=0;i<n;i++)
		{
			if(arr[i][0]==0)
			{
				for(int j=0;j<arr[i][1];j++)
				{
					f=f.append("(");
				}
			}
		}
		ArrayList<int[]> ar1=new ArrayList<>();
		ArrayList<int[]> ar2=new ArrayList<>();
		for(int i=0;i<n;i++)
		{
			if(arr[i][1]!=0 && arr[i][0]!=0 & arr[i][1]>=arr[i][0])
				ar1.add(new int[] {arr[i][0],arr[i][1]});
			else if(arr[i][1]!=0 && arr[i][0]!=0 & arr[i][1]<arr[i][0])
				ar2.add(new int[] {arr[i][0],arr[i][1]});
		}
		ar1.sort(new Comparator<int[]>() {
			public int compare(int[]a,int[]b)
			{
				if(a[0]>b[0])
					return 1;
				else if(a[0]<b[0])
					return -1;
				return 0;
			}
		});
		ar2.sort(new Comparator<int[]>() {
			public int compare(int[]a,int[]b)
			{
				if(a[1]>b[1])
					return -1;
				else if(a[1]<b[1])
					return 1;
				return 0;
			}
		});
		for(int i=0;i<ar1.size();i++)
		{
			for(int j=0;j<ar1.get(i)[0];j++)
			{
				f=f.append(")");
			}
			for(int j=0;j<ar1.get(i)[1];j++)
			{
				f=f.append("(");
			}
		}
		for(int i=0;i<ar2.size();i++)
		{
			for(int j=0;j<ar2.get(i)[0];j++)
			{
				f=f.append(")");
			}
			for(int j=0;j<ar2.get(i)[1];j++)
			{
				f=f.append("(");
			}
		}
		for(int i=0;i<n;i++)
		{
			if(arr[i][1]==0)
			{
				for(int j=0;j<arr[i][0];j++)
				{
					f.append(")");
				}
			}
		}
		Stack<Character> st=new Stack<>();
		int b=0;
		for(int i=0;i<f.length();i++)
		{
			if(f.charAt(i)=='(')
				st.push('(');
			else
			{
				if(st.empty())
				{
					b=1;
					break;
				}
				else
					st.pop();
			}
		}
		if(b==1 || st.size()!=0)
			System.out.println("No");
		else
			System.out.println("Yes");
	}
}
