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::input;
#[allow(unused_imports)]
use proconio::marker::*;

fn main() {
    input! {
        n: usize,
        s: [Bytes; n],
    }
    let mut l = 0;
    let mut m = vec![];
    let mut r = 0;
    for s in s {
        let mut v = vec![];
        for c in s {
            if c == b')' && v.last() == Some(&b'(') {
                v.pop();
            } else {
                v.push(c);
            }
        }
        match (v.first(), v.last()) {
            (Some(b'('), Some(b'(')) => l += v.len() as i64,
            (Some(b')'), Some(b'(')) => {
                let x = v.iter().filter(|&c| *c == b')').count() as i64;
                m.push((x, v.len() as i64 - x))
            }
            (Some(b')'), Some(b')')) => r += v.len() as i64,
            _ => {}
        }
    }
    m.sort_by_key(|m| m.0 - m.1);
    for m in m {
        if l >= m.0 {
            l -= m.0;
            l += m.1;
        } else {
            println!("No");
            return;
        }
    }
    if l == r {
        println!("Yes");
    } else {
        println!("No");
    }
}