#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int lowbit(int x) { return x & -x; }
int read() {
  int v = 0, f = 1;
  char c = getchar();
  while (c < 48 || 57 < c) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (48 <= c && c <= 57) v = v * 10 + c - 48, c = getchar();
  return v * f;
}
struct timer {
  double t;
  timer() { t = clock(); }
  ~timer() { printf("\nruntime %.3f s\n", get_time()); }
  double get_time() { return (clock() - t) / CLOCKS_PER_SEC; }
};
int a[123];
int main() {
  int n;
  string s;
  cin >> n >> s;
  int i = 0;
  while (i < n) {
    cout << s[i];
    i++;
    cout << s[i];
    i++;
    if (i == n - 1) {
      cout << s[i];
      i++;
    }
    if (i < n) cout << "-";
  }
  cout << endl;
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

Phone number in Berland is a sequence of n digits. Often, to make it easier to memorize the number, it is divided into groups of two or three digits. For example, the phone number 1198733 is easier to remember as 11-987-33. Your task is to find for a given phone number any of its divisions into groups of two or three digits.

Input

The first line contains integer n (2 ≤ n ≤ 100) — amount of digits in the phone number. The second line contains n digits — the phone number to divide into groups.

Output

Output any of divisions of the given phone number into groups of two or three digits. Separate groups by single character -. If the answer is not unique, output any.

Examples

Input

6
549871


Output

54-98-71

Input

7
1198733


Output

11-987-33

### Response

```cpp
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int lowbit(int x) { return x & -x; }
int read() {
  int v = 0, f = 1;
  char c = getchar();
  while (c < 48 || 57 < c) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (48 <= c && c <= 57) v = v * 10 + c - 48, c = getchar();
  return v * f;
}
struct timer {
  double t;
  timer() { t = clock(); }
  ~timer() { printf("\nruntime %.3f s\n", get_time()); }
  double get_time() { return (clock() - t) / CLOCKS_PER_SEC; }
};
int a[123];
int main() {
  int n;
  string s;
  cin >> n >> s;
  int i = 0;
  while (i < n) {
    cout << s[i];
    i++;
    cout << s[i];
    i++;
    if (i == n - 1) {
      cout << s[i];
      i++;
    }
    if (i < n) cout << "-";
  }
  cout << endl;
}

```