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++ (GCC 9.2.1) (judge verdict: Accepted)

#include "bits/stdc++.h"
#define rep(i,n) for(int i = 0; i < (n); ++i)
using namespace std;
typedef long long int ll;
typedef pair<int, int> P;
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 (a > b) { a = b; return 1; } return 0; }

const int mod = 1000000007;
struct mint{
	ll x;
	mint(ll 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{
		mint res(*this);
		return res+=a;
	}
	mint operator-(const mint a) const{
		mint res(*this);
		return res-=a;
	}
	mint operator*(const mint a) const{
		mint res(*this);
		return res*=a;
	}
	mint pow(ll t) const{
		if(!t) return 1;
		mint a = pow(t>>1);
		a *= a;
		if(t&1) a *= *this;
		return a;
	}
	mint inv() const{
		return pow(mod-2);
	}
	mint& operator/=(const mint a){
		return (*this) *= a.inv();
	}
	mint operator/(const mint a) const{
		mint res(*this);
		return res/=a;
	}
	bool operator==(const mint a){
		return x == a.x;
	}
};

// 2^n
mint f(int n){
	if(n == 0) return 1;
	mint x = f(n/2);
	x *= x;
	if(n%2 == 1) x *= 2;
	return x;
}

// nCa
mint choose(int n, int a){
	mint x = 1, y = 1;
	rep(i,a){
		x *= n-i;
		y *= i+1;
	}
	return x / y;
}

struct Combination{
	vector<mint> fact, ifact;
	Combination(int n):fact(n+1),ifact(n+1){
		assert(n < mod);
		fact[0] = 1;
		for(int i = 1; i <= n; i++) fact[i] = fact[i-1]*i;
		ifact[n] = fact[n].inv();
		for(int i = n; i >= 1; i--) ifact[i-1] = ifact[i]*i;
	}
	mint operator()(int n, int k){
		if(k < 0 || k > n) return 0;
		return fact[n]*ifact[k]*ifact[n-k];
	}
};

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