Phone numbers

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本文介绍了一种将电话号码按两位或三位分组的简单算法，以便于记忆。该算法根据电话号码的长度进行不同方式的分组，并通过示例说明了如何实现。

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Phone numbers

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.

Example

Input

6
549871

Output

54-98-71

Input

7
1198733

Output

11-987-33

题意：给一串长度为n的数字串，在两个或三个数字之间加-，便于记忆。

解题思路：可以将数字串都3个3个分，遇到余数不同的分情况讨论。

code：

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
int main(){
    int n,i,d,cnt = 0;
    char a[105];
    scanf("%d",&n);
    scanf("%s",a);
    if(n % 3 == 0){
        printf("%c%c%c",a[0],a[1],a[2]);
        for(i = 3; i < n; i += 3){
            printf("-%c%c%c",a[i],a[i+1],a[i+2]);
        }
        printf("\n");
    }
    else if(n % 3 == 1){
        printf("%c%c-%c%c",a[0],a[1],a[2],a[3]);
        for(i = 4; i < n; i += 3){
            printf("-%c%c%c",a[i],a[i+1],a[i+2]);
        }
        printf("\n");
    }
    else{
        printf("%c%c",a[0],a[1]);
        for(i = 2; i < n; i += 3){
            printf("-%c%c%c",a[i],a[i+1],a[i+2]);
        }
        printf("\n");
    }
    return 0;
}