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: Accepted)

use itertools::Itertools;
fn main() {
    proconio::input! {
        n: usize,
        ss: [proconio::marker::Chars; n],
    }
    let v = ss
        .iter()
        .map(|s| {
            let mut current = 0;
            let mut min = 0;
            s.iter().for_each(|c| {
                if *c == ')' {
                    current -= 1;
                    min = std::cmp::min(min, current);
                } else {
                    current += 1;
                }
            });
            (min, current)
        })
        .collect::<Vec<_>>();
    let positive = v.iter().filter(|(min, _current)| *min >= 0).collect_vec();
    let nega_posi = v
        .iter()
        .filter(|(min, current)| *min < 0 && *current >= 0)
        .sorted_by(|l, r| r.0.cmp(&l.0))
        .collect_vec();
    let negative = v
        .iter()
        .filter(|(min, current)| *min < 0 && *current < 0)
        .sorted_by(|l, r| (r.1 - r.0).cmp(&(l.1 - l.0)))
        .collect_vec();
    let mut current = 0;
    for (l, r) in positive.iter().chain(&nega_posi).chain(&negative) {
        if current + l < 0 {
            println!("No");
            return;
        }
        current += r;
    }
    if current == 0 {
        println!("Yes");
    } else {
        println!("No");
    }
}