CodeForces 219A k-String_a string is called a k-string if it can be represe-优快云博客

本文介绍了一种字符串重构问题：给定一个字符串s和整数k，如何通过重新排列s中的字母来构造一个k-string。文章提供了一个示例性的算法实现，并通过样例输入输出展示了该问题的解决过程。

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A string is called a k-string if it can be represented ask concatenated copies of some string. For example, the string "aabaabaabaab" is at the same time a 1-string, a 2-string and a 4-string, but it is not a 3-string, a 5-string, or a 6-string and so on. Obviously any string is a 1-string.

You are given a string s, consisting of lowercase English letters and a positive integerk. Your task is to reorder the letters in the strings in such a way that the resulting string is ak-string.

Input

The first input line contains integer k (1 ≤ k ≤ 1000). The second line containss, all characters in s are lowercase English letters. The string length s satisfies the inequality 1 ≤ |s| ≤ 1000, where|s| is the length of string s.

Output

Rearrange the letters in string s in such a way that the result is ak-string. Print the result on a single output line. If there are multiple solutions, print any of them.

If the solution doesn't exist, print "-1" (without quotes).

Sample test(s)

Input

2
aazz

Output

azaz

Input

3
abcabcabz

Output

-1

#include<cstdio>
#include<iostream>
using namespace std;

int main(){
    int k,i,j,kk,tt;
    char a[1002];
    int b[26];
    int flag;
    flag=0;
    scanf("%d",&k);
    scanf("%s",a);
    for(i=0;i<26;i++){
        b[i]=0;
    }
    for(i=0;a[i]!='\0';i++){
        b[a[i]-'a']++;
    }
    for(i=0;i<26;i++){
        if(b[i]%k!=0){
            flag=1;
            printf("-1\n");
            break;
        }
    }
    if(flag==0){
        for(tt=0;tt<k;tt++){
            for(i=0;i<26;i++){
                j=b[i]/k;
                for(kk=0;kk<j;kk++){
                    printf("%c",'a'+i);
                }
            }
        }
        printf("\n");
    }
    return 0;
}