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 range(i,a,b) for(int i = (a); i < (b); i++)
#define rep(i,b) for(int i = 0; i < (b); i++)
#define all(a) (a).begin(), (a).end()
#define show(x)  cerr << #x << " = " << (x) << endl;
#define int long long
using namespace std;

template <typename X, typename T>
auto vectors(X x, T a) {
	return vector<T>(x, a);
}

template <typename X, typename Y, typename Z, typename... Zs>
auto vectors(X x, Y y, Z z, Zs... zs) {
	auto cont = vectors(y, z, zs...);
	return vector<decltype(cont)>(x, cont);
}

template<typename T>
ostream& operator << (ostream& os, vector<T>& v){
	rep(i,v.size()){ os << v[i] << (i == v.size() - 1 ? "" : " "); } return os;
}
template<typename T>
istream& operator >> (istream& is, vector<T>& v){
	for(T& x: v){ is >> x; } return is;
}

const int M = 1000000007;

signed main(){
	int n;
	string s;
	cin >> n >> s;

	vector<int> num(26, 0);
	for(auto i : s){
		num[i - 'a']++;
	}


	int ans = 1;
	rep(i,num.size()){
		(ans *= num[i] + 1) %= M;
	}

	cout << ans - 1 << endl;
}