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, fastout};
use proconio::marker::Chars;
use std::cmp::Reverse;

#[derive(Debug)]
struct LR {
    l: isize,
    r: isize,
}

impl LR {
    fn new () -> Self {
        Self { l: 0, r: 0 }
    }
}

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

    let mut lrs1 = vec![];
    let mut lrs2 = vec![];

    for s in ss {
        let mut lr = LR::new();
        let mut num = 0;
        for c in s {
            match c {
                '(' => {
                    num += 1;
                },
                ')' => {
                    if num == 0 {
                        lr.l += 1;
                    } else {
                        num -= 1;
                    }
                }
                _ => {},
            }
        }
        if num > 0 {
            lr.r = num;
        }

        if lr.l <= lr.r {
            lrs1.push(lr)
        } else {
            lrs2.push(lr);
        }
    }



    lrs1.sort_by_key(|x| (x.l, x.r));
    lrs2.sort_by_key(|x| (Reverse(x.r), x.l));


    let mut num = 0;
    for lr in lrs1{
        num -= lr.l;
        if num < 0 {
            println!("No");
            return;
        }
        num += lr.r;
    }
    for lr in lrs2 {
        num -= lr.l;
        if num < 0 {
            println!("No");
            return;
        }
        num += lr.r;
    }
    
    if num == 0 {
        println!("Yes");
    } else {
        println!("No");
    }
}