Codeforces Round #352 (Div. 2)-B. Different is Good（模拟）

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 string s.

The second line contains the string s of length n 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".

题意：

要你找出不重复出现的子串，不符合改变字母使其符合。

思路：

这题很明显就是26个字母看是否够用，不够用就是-1.

AC代码：

#include<iostream>#include<cstdio>using namespace std;#define T 100000+50int main(){#ifdef zscfreopen("input.txt","r",stdin);#endifint n,m,i,j,k,ans;char s[T];int vis[30];while(~scanf("%d",&n)){ans = 0;k = 0;memset(vis,0,sizeof(vis));scanf("%s",s);for(i=0;s[i];++i){if(vis[s[i]-'a']>0)ans++;else k++;vis[s[i]-'a']++;}if(k+ans<=26)printf("%d\n",ans);elseprintf("-1\n");}return 0;}