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 all substrings 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
2
aa
Output
1
Input
4
koko
Output
2
Input
5
murat
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：长度为n的小写字母字符, 要求分解之后 不能有相同的字符串 ，问需要更改多少个字符，当无法实现时输出-1

题解：分析可知 当n>26 必然存在相同字符 输出-1

                     n<=26 处理 计算重复的数量  输出

 1 #include<bits/stdc++.h>
 2 #include<iostream>
 3 #include<cstring>
 4 #include<cstdio>
 5 #include<map>
 6 #include<queue>
 7 #include<stack>
 8 #define ll __int64
 9 #define pi acos(-1.0)
10 using namespace std;
11 char a[100005];
12 map<char,int>mp;
13 int n;
14 int main()
15 {
16    scanf("%d",&n);
17    mp.clear();
18    getchar();
19    for(int i=1;i<=n;i++)
20    {
21     scanf("%c",&a[i]);
22     mp[a[i]]++;
23    }
24    if(n>26)
25    {
26        cout<<"-1"<<endl;
27        return 0;
28    }
29    int ans=0;
30    for(int i='a';i<='z';i++)
31    {
32        if(mp[i]>1)
33        {
34            ans=ans+mp[i]-1;
35        }
36    }
37    cout<<ans<<endl;
38     return 0;
39 }