Problem:
Score : 600 points

Problem Statement

A bracket sequence is a string that is one of the following:

An empty string;

The concatenation of (, A, and ) in this order, for some bracket sequence A ;

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.

Sample Input 1

2
)
(()

Sample Output 1

Yes

Concatenating (() and ) in this order forms a bracket sequence.

Sample Input 2

2
)(
()

Sample Output 2

No

Sample Input 3

4
((()))
((((((
))))))
()()()

Sample Output 3

Yes

Sample Input 4

3
(((
)
)

Sample Output 4

No

Solution in Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.util.*;
public class Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		// 文字列の入力
		int n = sc.nextInt();

		boolean leftEnd = false;
		boolean rightEnd = false;
		int leftCount = 0;
		int rightCount = 0;
		for (int i = 0; i < n; i++) {
			String s = sc.next();
			while (s.contains("()")) {
				s = s.replace("()", "");
			}
			if (!s.equals("")) {
				int count = 0;
				while (count < s.length() && s.substring(count, count + 1).equals(")")) {
					count++;
				}
				int rightCloseCount = count;
				int leftCloseCount = s.length() - count;

				if (rightCloseCount == 0) {
					leftEnd = true;
				}
				if (leftCloseCount == 0) {
					rightEnd = true;
				}
				if (rightCloseCount < leftCloseCount) {
					leftCount += leftCloseCount - rightCloseCount;
				} else {
					rightCount += rightCloseCount - leftCloseCount;
				}
			}
		}

		// 出力
		System.out.println(leftEnd && rightEnd && leftCount == rightCount ? "Yes" : "No");
		sc.close();
	}
}