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)

use proconio::{fastout, input};

use proconio::marker::Chars;
use std::collections::BTreeSet;

#[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 ls = 0;
    let mut rs = 0;
    let mut lh = false;
    let mut rh = false;
    let mut lo = 0;
    let mut ro = 0;
    for &(l, r) in p.iter() {
        ls += l;
        rs += r;
        if l == 0 && r != 0 {
            lh = true;
            if r > lo {
                lo = r;
            }
        }
        if r == 0 && l != 0 {
            rh = true;
            if l > ro {
                ro = r;
            }
        }
    }
    if ls != rs {
        println!("No");
        return;
    }
    if !(lh && rh) {
        println!("No");
        return;
    }
    if p.iter().filter(|&(l, r)| *l <= lo).count() > 0
        && p.iter().filter(|&(l, r)| *r <= ro).count() > 0
    {
        println!("Yes");
    } else {
        println!("No");
    }
    // let mut h = BTreeSet::new();
    // h.insert((0, 0));
    // let pos = |x, y| if x > y { x - y } else { 0 };
    // for i in 0..n {
    //     let mut g = BTreeSet::new();
    //     let (l1, r1) = p[i];
    //     for &(l2, r2) in h.iter() {
    //         g.insert((l1 + pos(l2, r1), pos(r1, l2) + r2));
    //         g.insert((l2 + pos(l1, r2), pos(r2, l1) + r1));
    //     }
    //     h = g;
    // }
    // if h.contains(&(0, 0)) {
    //     println!("Yes");
    // } else {
    //     println!("No");
    // }
}