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 Lua (5.3.2) (judge verdict: Accepted)

local DBG = false
function dbgpr(...)
    if DBG then
        io.write("[dbg]")
        print(...)
    end
end

function dbgpr_t(tbl)
    if DBG then
        dbgpr(tbl)
        io.write("[dbg]")
        for i,v in ipairs(tbl) do
            io.write(i)
            io.write(":")
            io.write(tostring(v))
            io.write(" ")
        end
        print("")
    end
end

function dbgpr_tp(tbl)
    if DBG then
        dbgpr(tbl)
        io.write("[dbg]")
        for i,v in pairs(tbl) do
            io.write(i)
            io.write(":")
            io.write(tostring(v))
            io.write(" ")
        end
        print("")
    end
end

------------------------------------
local MOD = 1000000007
local AA, ZZ = 97, 122
local alpha_count = {}
for i=AA,ZZ,1 do
    alpha_count[i] = 0
end
local N = io.read("n", "l")
local S = io.read("l")
local s_tbl = {string.byte(S, 1, #S)}
for _,v in ipairs(s_tbl) do
    alpha_count[v] = alpha_count[v] + 1
end
dbgpr_tp(alpha_count)

local answer = 1
for i=AA,ZZ,1 do
    local place_cnt = alpha_count[i]
    if place_cnt == 0 then
    elseif place_cnt == 1 then
        answer = (answer * 2) % MOD
    else
        -- place_cnt >= 2
        local alphai_combs = 1 + place_cnt
        answer = (answer * alphai_combs) % MOD
    end
end
answer = (answer - 1) % MOD
print(answer)