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)

fn get_min_and_sum(s: &String) -> (i32, i32) {
    let mut now = 0;
    let mut mn = 0;
    for c in s.chars().as_str().chars() {
        if c == '(' {
            now += 1;
        } else {
            now -= 1;
        }
        mn = mn.min(now);
    }
    return (mn, now);
}

fn main() {
    proconio::input! {
        n: usize,
        strs: [String; n],
    }
    let mut pos = Vec::new();
    let mut neg = Vec::new();
    for s in strs.iter() {
        let mut tmp = get_min_and_sum(s);
        if tmp.1 >= 0 {
            pos.push(tmp);
        } else {
            tmp.0 -= tmp.1;
            tmp.1 = -tmp.1;
            neg.push(tmp);
        }
    }
    pos.sort();
    neg.sort();
    let mut ok = true;
    let mut now = 0;
    for (mn, sum) in pos {
        if now + mn < 0 {
            ok = false;
        }
        now += sum;
    }
    let mut now2 = 0;
    for (mn, sum) in neg {
        if now2 + mn < 0 {
            ok = false;
        }
        now2 += sum;
    }
    if ok && now == now2 {
        println!("Yes");
    } else {
        println!("No");
    }
}