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 proconio::{fastout, input};

use proconio::marker::Chars;

#[fastout]
fn main() {
    input! {
        n: usize,
        ss: [Chars; n],
    }
    let mut p = vec![];
    for s in ss {
        let mut r = 0;
        let mut m = 0;
        for c in s {
            match c {
                '(' => {
                    r -= 1;
                }
                ')' => {
                    r += 1;
                    if r > m {
                        m = r
                    }
                }
                _ => {}
            }
        }
        p.push((m, m - r));
    }
    let (mut first, mut second): (Vec<(i32, i32)>, Vec<(i32, i32)>) =
        p.iter().partition(|&(a, b)| b - a >= 0);
    first.sort_by_key(|&(a, _)| a);
    second.sort_by_key(|&(_, b)| std::cmp::Reverse(b));
    let mut r = 0;
    for &(a, b) in first.iter().chain(second.iter()) {
        r -= a;
        if r < 0 {
            println!("No");
            return;
        }
        r += b;
    }
    if r == 0 {
        println!("Yes");
    } else {
        println!("No");
    }
}