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 C (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<stdint.h>
#include<stdbool.h>
#include<tgmath.h>
#include<sys/queue.h>
#include<time.h>
#define ll long long
#define min(a, b) ((a) > (b) ? (b) : (a))
#define max(a, b) ((a) > (b) ? (a) : (b))
#define miin(a, b, c) min((a), min((b), (c)))
#define maax(a, b, c) max((a), max((b), (c)))
#define mod 1000000007

int n, a[1111111][2], nyaa=0, nyan[1111111];
char s[1111111];



int cmpare(const int *x, const int *y){
    int i = *x, j = *y;
    if(a[i][1]==0){
        if(a[j][1]==0){
            return a[j][0] - a[i][0];
        }
        return 1;
    }
    if(a[j][1]==0){
        return -1;
    }
    if(a[i][0] < a[j][0]) return -1;
    if(a[i][0] > a[j][0]) return 1;
    if(a[i][1] < a[j][1]) return 1;
    return -1;
}



main(){
    scanf("%d", &n);
    for(int i=0; i<n; i++){
        scanf("%s", s);
        int l = strlen(s);
        int b=0,c=0,d=0,e=0;
        for(int j=0; j<l; j++){
            if(s[j]==')')d++;
            else d--;
            if(s[l-1-j]=='(')e++;
            else e--;
            b = max(b,d);
            c = max(c,e);
        }
        a[i][0] = b;
        a[i][1] = c;
        nyan[i] = i;
    }
    qsort(nyan, n, 4, cmpare);
    if(a[nyan[0]][0] != 0 || a[nyan[n-1]][1] != 0){
        puts("No");
        return 0;
    }
    for(int i=0; i<n; i++){
        nyaa += a[nyan[i]][1] - a[nyan[i]][0];
        if(nyaa < 0)break;
    }
    puts(nyaa?"No":"Yes");
    return 0;
}