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

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 substringsof 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".

要求所有子集都各不相同，空集不用管，考虑极端情况集合中只有一个元素的时候，要保证所有子集各不相同，必须保证输入的字符串中每个元素各异，有重复的

要把重复的换成a - z中在字符串中没有出现过的，由此可见，若字符串长度大于26，根据鸽笼原理，那么肯定不能保证他们各异，因为a - z只有26个字母，第27个肯定至少跟

26个中的一个相同。在字符串小于等于26的时候去重就行了。

既然保证每个元素都互不相同，那么他们组成的更长的字符串肯定也互不相同

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <cstdlib>#include <stack>#include <queue>#include <string>using namespace std;const int maxn = 100000;char s[maxn + 10];int a[30];int main(){int n;scanf("%d", &n);scanf("%s", s);memset(a, 0, sizeof(a));for (int i = 0; i < n; i++) {a[s[i] - 'a']++;}if (n > 26) {puts("-1");}else {int ans = 0;for (int i = 0; i < 26; i++) {if (a[i])ans += (a[i] - 1);}printf("%d\n", ans);}return 0;}