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)

fun main(args: Array<String>) {
    val n = readLine()!!.toInt()
    val sList = (1..n).map { readLine()!! }
    val countLR = mutableListOf<Triple<Int, Int, Int>>()//index left right

    fun countLR(s: String): Pair<Int, Int> {
        var tl = 0
        var tr = 0
        for (c in s) {
            if (c == '(') {
                tr++
            } else {
                if (tr > 0) tr--
                else tl++
            }
        }
        return Pair(tl, tr)
    }
    for (i in 0 until n) {
        val (l, r) = countLR(sList[i])
        countLR.add(Triple(i, l, r))
    }
    countLR.sortBy { it.second }
    val temp = mutableListOf<String>()
    for (i in 0 until n) {
        val (index, l, r) = countLR[i]
        if (l == 0 && r != 0) temp.add(sList[index])
    }
    for (i in 0 until n) {
        val (index, l, r) = countLR[i]
        if (l != 0 && r != 0) temp.add(sList[index])
    }
    for (i in 0 until n) {
        val (index, l, r) = countLR[i]
        if (l != 0 && r == 0) temp.add(sList[index])
    }
    val anss = temp.joinToString("")
    println(anss)
    val (l, r) = countLR(anss)
    if (l == 0 && r == 0) {
        println("Yes")
    } else {
        println("No")
    }
}