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::*;
use proconio::marker::Chars;

// 中のカッコは消し飛ばす
// )))
// (((
// ))(((
// 無
// の4つになる
// 括弧の数が一致するのは必須
// (( -> ))(( -> ))
// の順で使って大丈夫か判定?
// mahoutukaippoi

fn run() {
    input! {
        n: usize,
        s: [Chars; n],
    }
    let mut a = vec![];
    let mut b = vec![];
    for s in s {
        let mut pos = 0i64;
        let mut min = 0i64;
        for c in s {
            if c == '(' {
                pos += 1;
            } else {
                pos -= 1;
            }
            min = std::cmp::min(min, pos);
        }
        let down = -min;
        let up = pos - min;
        if up >= down {
            a.push((down, up));
        } else {
            b.push((down, up));
        }
    }
    a.sort_by_key(|a| a.0);
    b.sort_by_key(|b| b.1);
    let mut s = 0;
    let mut ans = true;
    for (d, u) in a.into_iter() {
        s -= d;
        ans &= s >= 0;
        s += u;
    }
    for (d, u) in b.into_iter().rev() {
        s -= d;
        ans &= s >= 0;
        s += u;
    }
    ans &= s == 0;
    let out = if ans {
        "Yes"
    } else {
        "No"
    };
    println!("{}", out);
}

fn main() {
    run();
}