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: Accepted)

#include<stdio.h>
int R=1,C=1,H[2000010],N[2000010];
//評価関数（いまはMIN）
int hyouka(int a,int b){
  if(C<b)return 1;
  if(C<a||b==0)return 0;
  return N[H[a]]>N[H[b]]?1:0;
}
//挿入関数
void hin(int a){
  int i=C++;
  for(N[H[0]=R]=a;hyouka(0,i/2);i/=2)H[i]=H[i/2];
  H[i]=R++;
}
//取り出す関数
int hout(){
  int rt=H[1],i,j=2,k=H[--C];
  for(i=1;hyouka(i,C);i=j)H[i]=H[j=i*2+1-hyouka(i*2,i*2+1)];
  H[j/2]=k;
  return rt;
}
int n,i,j,a,b,d[1000010],e[1000010],f[1000010],min;
int main(){
  char s[1000010];
  scanf("%d",&n);
  for(i=0;i<n;i++){
    scanf("%s",s);
    for(j=a=min=0;s[j];j++){
      if(s[j]=='(')a++;
      else         a--;
      if(min>a)min=a;
    }
    d[i]=min;
    e[i]=a;
  }
  for(i=0;i<n;i++){
    if(e[i]>=0){
      f[R]=e[i];
      hin(d[i]);
    }
  }
  a=0;
  while(C-1){
    i=hout();
    if(a+N[i]<0)goto END;
    a+=f[i];
  }
  R=C=1;
  for(i=0;i<n;i++){
    if(e[i]<0){
      f[R]=-e[i];
      hin(d[i]-e[i]);
    }
  }
  b=0;
  while(C-1){
    i=hout();
    if(b+N[i]<0)goto END;
    b+=f[i];
  }
  if(a-b)goto END;
  printf("Yes\n");
  return 0;
 END:;
  printf("No\n");
  return 0;
}