Problem: A sequence of square brackets is regular if by inserting symbols "+" and "1" into it, you can get a regular mathematical expression from it. For example, sequences "[[]][]", "[]" and "[[][[]]]" — are regular, at the same time "][", "[[]" and "[[]]][" — are irregular. 

Draw the given sequence using a minimalistic pseudographics in the strip of the lowest possible height — use symbols '+', '-' and '|'. For example, the sequence "[[][]][]" should be represented as: 
    
    
      
    +-        -++- -+      
    |+- -++- -+||   |  
    ||   ||   |||   |  
    |+- -++- -+||   |  
    +-        -++- -+  
    

Each bracket should be represented with the hepl of one or more symbols '|' (the vertical part) and symbols '+' and '-' as on the example which is given above.

Brackets should be drawn without spaces one by one, only dividing pairs of consecutive pairwise brackets with a single-space bar (so that the two brackets do not visually merge into one symbol). The image should have the minimum possible height. 

The enclosed bracket is always smaller than the surrounding bracket, but each bracket separately strives to maximize the height of the image. So the pair of final brackets in the example above occupies the entire height of the image.

Study carefully the examples below, they adequately explain the condition of the problem. Pay attention that in this problem the answer (the image) is unique. 

Input

The first line contains an even integer n (2 ≤ n ≤ 100) — the length of the sequence of brackets.

The second line contains the sequence of brackets — these are n symbols "[" and "]". It is guaranteed that the given sequence of brackets is regular. 

Output

Print the drawn bracket sequence in the format which is given in the condition. Don't print extra (unnecessary) spaces. 

Examples

Input

8
[[][]][]


Output

+-        -++- -+
|+- -++- -+||   |
||   ||   |||   |
|+- -++- -+||   |
+-        -++- -+


Input

6
[[[]]]


Output

+-     -+
|+-   -+|
||+- -+||
|||   |||
||+- -+||
|+-   -+|
+-     -+


Input

6
[[][]]


Output

+-        -+
|+- -++- -+|
||   ||   ||
|+- -++- -+|
+-        -+


Input

2
[]


Output

+- -+
|   |
+- -+


Input

4
[][]


Output

+- -++- -+
|   ||   |
+- -++- -+

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class T, class U>
inline void relax_min(T &a, U b) {
  a = min(a, b);
}
template <class T, class U>
inline void relax_max(T &a, U b) {
  a = max(a, b);
}
template <typename T>
inline bool updmax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
template <typename T>
inline bool updmin(T &a, const T &b) {
  return a > b ? a = b, 1 : 0;
}
template <class T, class U>
istream &operator>>(istream &in, pair<T, U> &a) {
  in >> a.first >> a.second;
  return in;
}
template <class T, class U>
ostream &operator<<(ostream &out, const pair<T, U> &a) {
  out << a.first << ' ' << a.second;
  return out;
}
template <class T>
istream &operator>>(istream &in, vector<T> &a) {
  for (decltype(0)((i)) = (0); ((i)) < ((((int)(a).size()))); ++((i)))
    in >> a[i];
  return in;
}
template <class T>
ostream &operator<<(ostream &out, const vector<T> &a) {
  for (decltype(0)((i)) = (0); ((i)) < ((((int)(a).size()))); ++((i)))
    out << a[i] << ' ';
  return out;
}
int next_int() {
  int(tmp);
  scanf("%d", &(tmp));
  return tmp;
}
long long next_ll() {
  long long(tmp);
  scanf("%lld", &(tmp));
  return tmp;
}
long long modPow(long long a, long long b, long long MOD) {
  long long ret = 1;
  for (; b; b >>= 1) {
    if (b & 1)
      ret = ((((((long long)(ret) * (long long)(a))) % ((MOD))) + ((MOD))) %
             ((MOD)));
    a = ((((((long long)(a) * (long long)(a))) % ((MOD))) + ((MOD))) % ((MOD)));
  }
  return ret;
}
const long long EXACTLY_ONE_HUNDRED = 100;
const long long EXACTLY_ONE_THOUSAND = 1000;
const long long EXACTLY_ONE_MILLION = 1000 * 1000;
const long long EXACTLY_ONE_BILLION =
    EXACTLY_ONE_THOUSAND * EXACTLY_ONE_MILLION;
const int MAXN1 = (int)(EXACTLY_ONE_HUNDRED * EXACTLY_ONE_THOUSAND) + 3;
const int MAXN2 = (int)(2 * EXACTLY_ONE_HUNDRED * EXACTLY_ONE_THOUSAND) + 3;
const int MAXN3 = (int)(3 * EXACTLY_ONE_HUNDRED * EXACTLY_ONE_THOUSAND) + 3;
const int MAXN4 = (int)(4 * EXACTLY_ONE_HUNDRED * EXACTLY_ONE_THOUSAND) + 3;
const int MAXN5 = (int)(5 * EXACTLY_ONE_HUNDRED * EXACTLY_ONE_THOUSAND) + 3;
const int MAXN = MAXN3;
const int INF = (int)EXACTLY_ONE_BILLION + 7;
const int INF2 = 1 << 30;
const long long INFLL = 4LL * EXACTLY_ONE_BILLION * EXACTLY_ONE_BILLION + 7;
const long long INFLL2 = 1LL << 60;
const long double Pi = acos(-1.0);
const long double EPS = 1e-7;
const int MOD = (int)EXACTLY_ONE_BILLION + 7;
char ans[500][2000];
int main() {
  int(n);
  scanf("%d", &(n));
  string brack;
  cin >> brack;
  int max_size = 0;
  int delta = 0;
  for (decltype(0)((i)) = (0); ((i)) < ((n)); ++((i))) {
    if (brack[i] == '[')
      ++delta;
    else
      --delta;
    updmax(max_size, delta);
  }
  int rows = 2 * max_size + 1;
  for (decltype(0)((i)) = (0); ((i)) < ((rows)); ++((i))) {
    memset((ans[i]), (' '), sizeof(ans[i]));
  }
  int cur_col = 0;
  delta = 0;
  for (decltype(0)((i)) = (0); ((i)) < ((n)); ++((i))) {
    if (i && brack[i - 1] == '[' && brack[i] == ']') {
      ++cur_col;
      ++cur_col;
    }
    if (i && brack[i - 1] == ']' && brack[i] == '[') ++cur_col;
    if (brack[i] == '[') {
      int height = rows - delta - 1;
      int cur_size = 2 * (max_size - delta) - 1;
      ans[delta][cur_col] = ans[height][cur_col] = '+';
      ans[delta][cur_col + 1] = ans[height][cur_col + 1] = '-';
      for (decltype(delta + 1)(j) = (delta + 1); (j) < (cur_size + delta + 1);
           ++(j)) {
        ans[j][cur_col] = '|';
      }
      ++delta;
    } else {
      --delta;
      int height = rows - delta - 1;
      int cur_size = 2 * (max_size - delta) - 1;
      ans[delta][cur_col] = ans[height][cur_col] = '-';
      ans[delta][cur_col + 1] = ans[height][cur_col + 1] = '+';
      for (decltype(delta + 1)(j) = (delta + 1); (j) < (cur_size + delta + 1);
           ++(j)) {
        ans[j][cur_col + 1] = '|';
      }
    }
    ++cur_col;
  }
  for (decltype(0)((i)) = (0); ((i)) < ((rows)); ++((i))) {
    ans[i][cur_col + 1] = 0;
    puts(ans[i]);
  }
  return 0;
}