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 Kotlin (1.3.71) (judge verdict: Wrong Answer)

import java.util.*
import kotlin.collections.ArrayList
import kotlin.math.max
import kotlin.random.Random

fun main() {
    val sc = Scanner(System.`in`)
    var n = sc.nextInt()
    var arr = Array(n){""}
    var m = Array(n){ArrayList<Int>()}

    fun createP(n:Int): String {
        val sb = StringBuilder()
        val c = arrayOf('(',')')
        for(i in 0 until n) {
            sb.append(c[Random.nextInt(0, 100) % 2])
        }
        return sb.toString()
    }
    fun createRand() {
        n = (7000).toInt()
        arr = Array(n){""}
        for(i in 0 until n) {
            val len = Random.nextInt(1, 2200)
            arr[i] = createP(len)
        }
        m = Array(n){ArrayList<Int>()}
    }

    for(i in 0 until n) {
        arr[i] = sc.next()
        arr[i] = arr[i].trim()
    }

//    createRand()

    for(i in 0 until n) {
        var k = 0
        val l = ArrayList<Int>()
        for(c in arr[i]) {
            if(c == '(') {
                if(k < 0) {
                    l.add(k)
                    k = 0
                }
                ++k
            } else if(c==')'){
                --k
            }
        }
        if(k != 0) {
            l.add(k)
        }
        m[i] = l
    }

    fun ex(d:Int): Int {
        if(d > 0) return 1
        return max(-1, d)
    }

    m.sortWith(kotlin.Comparator { o1, o2 ->
        if(o1.size == o2.size) {
            for(i in o1.indices) {
                val l = ex(o1[i])
                val r = ex(o2[i])
                if(l == r) continue
                return@Comparator r.compareTo(l)
            }
            return@Comparator 0
        }
        for(i in 0 until max(o1.size, o2.size)) {
            val l = if(i >= o1.size) {
                0
            } else {
                ex(o1[i])
            }
            val r = if(i >= o2.size) {
                0
            } else {
                ex(o2[i])
            }
            if(l == r) continue
            return@Comparator r.compareTo(l)
        }
        return@Comparator 0
    })

    val ans = ArrayList<Int>()
    m.forEach {x ->
        x.forEach { ans.add(it) }
    }

    var acc = 0L
    for(d in ans) {
        acc += d
        if(acc < 0L) {
            break
        }
    }

    if(acc == 0L) {
        print("Yes")
    } else {
        print("No")
    }


}