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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define rep(i,n) for(int i = 0; i < (int)(n); i++)
using namespace std;
using ll = long long;
using P = pair<int, int>;
using vi = vector<int>;
using vc = vector<char>;
using vs = vector<string>;
using vll = vector<long long>;
using vvi = vector<vector<int>>;
using vvc = vector<vector<char>>;
using vvll = vector<vector<long long>>;
template<class T> inline bool chmax(T &a, T b) { if (a<b) {a=b; return 1;} return 0;}
template<class T> inline bool chmin(T &a, T b) { if (b<a) {a=b; return 1;} return 0;}

const int mod = 1000000007;
// const int mod = 998244353;
struct mint {
    long long x;
    mint(long long x=0) : x((x % mod + mod) % mod) {}
    mint operator-() const { return mint(-x);}
    mint& operator+=(const mint a) {
        if ((x += a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator-=(const mint a) {
        if ((x += mod - a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}
    mint operator+(const mint a) const { return mint(*this) += a;}
    mint operator-(const mint a) const { return mint(*this) -= a;}
    mint operator*(const mint a) const { return mint(*this) *= a;}
    mint pow(long long t) const {
        if (!t) return 1;
        mint a = pow(t>>1);
        a *= a;
        if (t&1) a *= *this;
        return a;
    }
    
    // for prime mod
    mint inv() const { return pow(mod-2);}
    mint& operator/=(const mint a) { return *this *= a.inv();}
    mint operator/(const mint a) const { return mint(*this) /= a;}
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    
    int n;
    string s;
    cin >> n >> s;
    map<char, int> mp;
    rep(i, n) mp[s[i]]++;
    mint ans = 1;
    for (auto p : mp) {
        ans *= p.second + 1;
    }
    ans -= 1;
    cout << ans << endl;
}