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;

fn check(arr : &Vec<(i32, i32)>) -> i32 {
    let mut res = 0;
    for &a in arr {
        res -= a.1;
        if res < 0 { return -1; }
        res += a.0;
    }
    return res;
}

fn main() {
    input! {
        n : usize,
        s : [String; n]
    };
    let mut up = Vec::new();
    let mut down = Vec::new();
    for c in s {
        let mut p = (0, 0);
        for &v in c.as_bytes() {
            if v == b'(' { p.0 += 1 }
            else {
                if p.0 > 0 { p.0 -= 1 }
                else { p.1 += 1 }
            }
        }
        if p.0 >= p.1 {
            up.push(p)
        } else {
            down.push((p.1, p.0))
        }
    }
    up.sort_by_key(|x| x.1);
    down.sort_by_key(|x| x.1);
    let a = check(&up);
    let b = check(&down);
    println!("{}", if a >= 0 && b >= 0 && a == b { "Yes" } else { "No" });
}