Problem:
Score : 200 points

Problem Statement

You are given a string S of length N.
Among its subsequences, count the ones such that all characters are different, modulo 10^9+7. Two subsequences are considered different if their characters come from different positions in the string, even if they are the same as strings.

Here, a subsequence of a string is a concatenation of one or more characters from the string without changing the order.

Constraints

1 \leq N \leq 100000

S consists of lowercase English letters.

|S|=N

Input

Input is given from Standard Input in the following format:

N
S

Output

Print the number of the subsequences such that all characters are different, modulo 10^9+7.

Sample Input 1

4
abcd

Sample Output 1

15

Since all characters in S itself are different, all its subsequences satisfy the condition.

Sample Input 2

3
baa

Sample Output 2

5

The answer is five: b, two occurrences of a, two occurrences of ba. Note that we do not count baa, since it contains two as.

Sample Input 3

5
abcab

Sample Output 3

17

Solution in Rust (1.42.0) (judge verdict: Wrong Answer)

#[allow(unused_imports)]
use proconio::marker::{Bytes, Chars, Usize1};
use proconio::{fastout, input};
use std::collections::HashMap;
#[fastout]
fn main() {
    input! {
        _n: usize,
        s: Chars,
    }
    let mut map: HashMap<char, usize> = HashMap::new();
    let mut ans = 0;
    let mut first = true;
    let waru = 1_000_000_000 + 7;
    for i in s.iter() {
        let counter = map.entry(*i).or_insert(0);
        *counter += 1;
        if *counter >= 2 {
            if first {
                ans += 2u32.pow(map.len() as u32) - 1 % waru;
                first = false;
            }
            // 2の(個数が1個の文字の数)乗 * (複数の文字の個数)
            let mut count_char = 0;
            let mut count_chars = 1;
            for i in map.values() {
                match i {
                    1 => count_char += 1,
                    _ => count_chars *= (*i as u32) % waru,
                }
            }
            ans += ((2u32.pow(count_char as u32) - 1) * count_chars as u32) % waru;
        }
    }
    if first {
        ans = 2u32.pow(map.len() as u32) - 1 % waru;
    }
    println!("{}", ans);
}