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 Rust (1.42.0) (judge verdict: Wrong Answer)

use proconio::input;

fn main() {
    input! {
        n:usize,
        ss:[proconio::marker::Chars;n]
    };

    let mut t:Vec<(i32,i32)> = ss.iter().map(|s|{
        let mut sum = 0;
        let mut minv = 0;
        for c in s {
            sum+= match c {
                '(' => 1,
                _ => -1,
            };
            minv = std::cmp::min(minv,sum);
        }
        (minv,sum)
    }).collect();

    t.sort_by_key(|(minv,sum)|
        (-sum.signum(),-minv)
    );
    let t = t;
    let mut sum = 0;
    for (minv,plus) in t {
        if sum+minv<0{
            println!("No");
            return;
        }
        sum+=plus;
    }
    if sum!=0{
        println!("No");
    }
    else {
        println!("Yes");
    }

    

    


}