Problem:
Score : 400 points

Problem Statement

You are given a string S of length N consisting of ( and ). Your task is to insert some number of ( and ) into S to obtain a correct bracket sequence.

Here, a correct bracket sequence is defined as follows:

() is a correct bracket sequence.

If X is a correct bracket sequence, the concatenation of (, X and ) in this order is also a correct bracket sequence.

If X and Y are correct bracket sequences, the concatenation of X and Y in this order is also a correct bracket sequence.

Every correct bracket sequence can be derived from the rules above.

Find the shortest correct bracket sequence that can be obtained. If there is more than one such sequence, find the lexicographically smallest one.

Constraints

The length of S is N.

1 ≤ N ≤ 100

S consists of ( and ).

Input

Input is given from Standard Input in the following format:

N
S

Output

Print the lexicographically smallest string among the shortest correct bracket sequences that can be obtained by inserting some number of ( and ) into S.

Sample Input 1

3
())

Sample Output 1

(())

Sample Input 2

6
)))())

Sample Output 2

(((()))())

Sample Input 3

8
))))((((

Sample Output 3

(((())))(((())))

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>

/*リスト一式*/
/*NDDTで置換する型を変更してリストのデータを指定*/
typedef char NDDT;
typedef struct node{
  NDDT data;
  struct node *next;
  struct node *prev;
} NODE;
typedef struct list{
  int size;
  NODE *front;
  NODE *back;
} LIST;
NODE *new_node(NDDT d){
  NODE *ret = (NODE *)malloc(sizeof(NODE));
  ret->data = d;
  ret->next = NULL;
  ret->prev = NULL;
  return ret;
}
void new_list(LIST *list){
  list->back = NULL;
  list->front = NULL;
  list->size = 0;
}
void push_back(NDDT d, LIST *list){
  NODE *new = new_node(d);
  if(list->size == 0){
    list->front = list->back = new;
  }else{
    list->back->next = new;
    new->prev = list->back;
    list->back = new;
  }
  list->size++;
}
void push_front(NDDT d, LIST *list){
  NODE *new = new_node(d);
  if(list->size == 0){
    list->front = list->back = new;
  }else{
    list->front->prev = new;
    new->next = list->front;
    list->front = new;
  }
  list->size++;
}
int pop_back(LIST *list){
  if(list == NULL || list->size == 0){
    return 1;
  }
  if(list->size == 1){
    free(list->back);
    list->front = list->back = NULL;
    list->size = 0;
    return 0;
  }
  NODE *bk = list->back;
  NODE *nbk = bk->prev;
  free(bk);
  nbk->next = NULL;
  list->back = nbk;
  list->size--;
  return 0;
}
int pop_front(LIST *list){
  if(list == NULL || list->size == 0){
    return 1;
  }
  if(list->size == 1){
    free(list->back);
    list->front = list->back = NULL;
    list->size = 0;
    return 0;
  }
  NODE *fr = list->front;
  NODE *nfr = fr->next;
  free(fr);
  nfr->prev = NULL;
  list->front = nfr;
  list->size--;
  return 0;
}
/*ここまでリスト一式*/


int main(void){
  LIST list;
  new_list(&list);
  int n;
  char s[105];
  scanf("%d", &n);
  scanf("%s", s);
  int co = 0;
  for(int i = 0; i < n; i++){
    char now = s[i];
    if(now == '(') push_back(now, &list);
    else if(list.size != 0){
      pop_back(&list);
    }else{
      co++;
    }
  }

  for(int i = 0; i < co; i++){
    printf("(");
  }
  for(int i = 0; i < n; i++){
    printf("%c", s[i]);
  }
  for(int i = 0; i < list.size; i++){
    printf(")");
  }
  printf("\n");
  while(list.size != 0){
    pop_back(&list);
  }
  return 0;
}