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.math.*;
import java.util.*;


public class Main{
    static long MOD = 998244353L;
    static long [] fac;

    public static void main(String[] args) {
        MyScanner sc = new MyScanner();
        out = new PrintWriter(new BufferedOutputStream(System.out));
        // Start writing your solution here. -------------------------------------
        //int t = sc.nextInt();
        int t = 1;
        while(t-- > 0)
        {
            int n = sc.nextInt();
            int cur = 0;
            boolean valid = true;
            int [][] a = new int[n][3];
            for(int i = 0; i < n; i++)
            {
                String s = sc.nextLine();
                int res = 0, min = 0;
                for(char c : s.toCharArray())
                {
                    if(c == '(') res++;
                    else res--;
                    min = Math.min(min, res);

                }
                a[i][0] = min;
                a[i][1] = res;
                a[i][2] = min - res;
            }
            Arrays.sort(a, (m1, m2) -> m1[1] * m2[1] <= 0 ? m2[1] - m1[1] : (m1[1] + m2[1] > 0 ? m2[0] - m1[0] : m1[2] - m2[2]));
            for(int [] m : a)
            {
                if(cur + m[0] < 0) valid = false;
                else cur += m[1];
            }
            if(cur != 0) valid = false;
            out.println(valid ? "Yes" : "No");
        }
        out.close();
    }
    public static long C(int n, int m)
    {
        if(m == 0 || m == n) return 1l;
        if(m > n || m < 0) return 0l;
        long res = fac[n] * quickPOW((fac[m] * fac[n - m]) % MOD, MOD - 2) % MOD;

        return res;
    }
    public static long quickPOW(long n, long m)
    {
        long ans = 1l;
        while(m > 0)
        {
            if(m % 2 == 1)
                ans = (ans * n) % MOD;
            n = (n * n) % MOD;
            m >>= 1;
        }
        return ans;
    }
    public static int gcd(int a, int b)
    {
        if(a % b == 0) return b;
        return gcd(b, a % b);
    }
    public static long solve(long x, long[] res){
        int n = res.length;
        long a = x / n;
        int b = (int)(x % n);
        return res[n - 1] * a + res[b];
    }
    //-----------PrintWriter for faster output---------------------------------
    public static PrintWriter out;
    //-----------MyScanner class for faster input----------
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int nextInt() {
            return Integer.parseInt(next());
        }
        long nextLong() {
            return Long.parseLong(next());
        }
        double nextDouble() {
            return Double.parseDouble(next());
        }
        String nextLine(){
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
    //--------------------------------------------------------
}