"QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.

Now Diamond has given Bort a string consisting of only uppercase English letters of length n. There is a great number of "QAQ" in the string (Diamond is so cute!).

<image> illustration by 猫屋 https://twitter.com/nekoyaliu

Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don't have to be consecutive, but the order of letters should be exact.

Input

The only line contains a string of length n (1 ≤ n ≤ 100). It's guaranteed that the string only contains uppercase English letters.

Output

Print a single integer — the number of subsequences "QAQ" in the string.

Examples

Input

QAQAQYSYIOIWIN


Output

4


Input

QAQQQZZYNOIWIN


Output

3

Note

In the first example there are 4 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN".
'''input
QAQQQZZYNOIWIN
'''
'''
Author : dhanyaabhirami
Great things never come from comfort zones
'''
from collections import defaultdict as dd
from collections import Counter as ccd
from itertools import  permutations as pp
from itertools import combinations as cc
from random import randint as rd
from bisect import bisect_left as bl
from bisect import bisect_right as br
import heapq as hq
import math 

mod=pow(10,9) +7
def inp(flag=0):
    if flag==0:
        return list(map(int,input().strip().split(' ')))
    else:
        return int(input())

# Code credits
# assert(debug()==true)
# for _ in range(int(input())):

def count_recursion(i,j):
	if j==len(target):
		return 1
	if i==len(source):
		return 0
	if source[i]==target[j]:
		return count_recursion(i+1,j+1)+count_recursion(i+1,j)
	else:
		return count_recursion(i+1,j)

source = input().strip()
target = 'QAQ'
print(count_recursion(0,0))
