Codeforces 672B Different is Good【水题】

B. Different is Good

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

A wise man told Kerem "Different is good" once, so Kerem wants all things in his life to be different.

Kerem recently got a string s consisting of lowercase English letters. Since Kerem likes it when things are different, he wants allsubstrings of his string s to be distinct. Substring is a string formed by some number of consecutive characters of the string. For example, string "aba" has substrings "" (empty substring), "a", "b", "a", "ab", "ba", "aba".

If string s has at least two equal substrings then Kerem will change characters at some positions to some other lowercase English letters. Changing characters is a very tiring job, so Kerem want to perform as few changes as possible.

Your task is to find the minimum number of changes needed to make all the substrings of the given string distinct, or determine that it is impossible.

Input

The first line of the input contains an integer n (1 ≤ n ≤ 100 000) — the length of the strings.

The second line contains the string s of lengthn consisting of only lowercase English letters.

Output

If it's impossible to change the string s such that all its substring are distinct print-1. Otherwise print the minimum required number of changes.

Examples

Input

2aa

Output

1

Input

4koko

Output

2

Input

5murat

Output

0

Note

In the first sample one of the possible solutions is to change the first character to 'b'.

In the second sample, one may change the first character to 'a' and second character to 'b', so the string becomes "abko".

题目大意：

给你一个长度为N的由小写字母组成的字符串，问你能否找到一种方案，使得这个字符串的所有子串都不相等。

如果有，输出最少修改的字符个数，否则输出-1.

思路:

对于一个字符串来讲，所有子串必然包括每个位子的字符作为单个子串，那么其实问题就是让你改变最少字符个数，使得每个字符都不同。

那么接下来找好姿势随便贪心即可。

Ac代码：

#include<stdio.h>#include<string.h>using namespace std;char a[100070];int vis[300];int main(){    int n;    while(~scanf("%d",&n))    {        memset(vis,0,sizeof(vis));        scanf("%s",a);        int ok=1;        int output=0;        for(int i=0;i<n;i++)        {            vis[a[i]]++;        }        for(int i=0;i<n;i++)        {            if(vis[a[i]]>1)            {                int flag=0;                for(int j='a';j<='z';j++)                {                    if(vis[j]==0)                    {                        output++;                        flag=1;                        vis[j]=1;                        vis[a[i]]--;                        break;                    }                }                if(flag==0)ok=0;            }        }        for(int i=0;i<n;i++)if(vis[a[i]]>1)ok=0;        if(ok==0)printf("-1\n");        else        printf("%d\n",output);    }}