import sys
import heapq
import re
from itertools import permutations
from bisect import bisect_left, bisect_right
from collections import Counter, deque
from fractions import gcd
from math import factorial, sqrt, ceil
from functools import lru_cache, reduce
INF = 1 << 60
MOD = 1000000007
sys.setrecursionlimit(10 ** 7)
 
# UnionFind
class UnionFind():
    def __init__(self, n):
        self.n = n
        self.parents = [-1] * n
 
    def find(self, x):
        if self.parents[x] < 0:
            return x
        else:
            self.parents[x] = self.find(self.parents[x])
            return self.parents[x]
 
    def union(self, x, y):
        x = self.find(x)
        y = self.find(y)
 
        if x == y:
            return
 
        if self.parents[x] > self.parents[y]:
            x, y = y, x
 
        self.parents[x] += self.parents[y]
        self.parents[y] = x
 
    def size(self, x):
        return -self.parents[self.find(x)]
 
    def same(self, x, y):
        return self.find(x) == self.find(y)
 
    def members(self, x):
        root = self.find(x)
        return [i for i in range(self.n) if self.find(i) == root]
 
    def roots(self):
        return [i for i, x in enumerate(self.parents) if x < 0]
 
    def group_count(self):
        return len(self.roots())
 
    def all_group_members(self):
        return {r: self.members(r) for r in self.roots()}
 
    def __str__(self):
        return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())
 
def is_prime(n):
    if n == 1:
        return False
    for i in range(2,int(n**0.5)+1):
        if n % i == 0:
            return False
    return True
 
 
# ワーシャルフロイド (任意の2頂点の対に対して最短経路を求める)
# 計算量n^3 (nは頂点の数)
def warshall_floyd(d, n):
    #d[i][j]: iからjへの最短距離
    for k in range(n):
        for i in range(n):
            for j in range(n):
                d[i][j] = min(d[i][j],d[i][k] + d[k][j])
    return d
 
 
# ダイクストラ
def dijkstra_heap(s, edge, n):
    #始点sから各頂点への最短距離
    d = [10**20] * n
    used = [True] * n #True:未確定
    d[s] = 0
    used[s] = False
    edgelist = []
    for a,b in edge[s]:
        heapq.heappush(edgelist,a*(10**6)+b)
    while len(edgelist):
        minedge = heapq.heappop(edgelist)
        #まだ使われてない頂点の中から最小の距離のものを探す
        if not used[minedge%(10**6)]:
            continue
        v = minedge%(10**6)
        d[v] = minedge//(10**6)
        used[v] = False
        for e in edge[v]:
            if used[e[1]]:
                heapq.heappush(edgelist,(e[0]+d[v])*(10**6)+e[1])
    return d
 
# 素因数分解
def factorization(n):
    arr = []
    temp = n
    for i in range(2, int(-(-n**0.5//1))+1):
        if temp%i==0:
            cnt=0
            while temp%i==0:
                cnt+=1
                temp //= i
            arr.append([i, cnt])
 
    if temp!=1:
        arr.append([temp, 1])
 
    if arr==[]:
        arr.append([n, 1])
 
    return arr
 
# 2数の最小公倍数
def lcm(x, y):
    return (x * y) // gcd(x, y)
 
# リストの要素の最小公倍数
def lcm_list(numbers):
    return reduce(lcm, numbers, 1)
 
# リストの要素の最大公約数
def gcd_list(numbers):
    return reduce(gcd, numbers)
 
# 素数判定
 
 
 
# limit以下の素数を列挙
def eratosthenes(limit):
    A = [i for i in range(2, limit+1)]
    P = []
 
    while True:
        prime = min(A)
        
        if prime > sqrt(limit):
            break
            
        P.append(prime)
            
        i = 0
        while i < len(A):
            if A[i] % prime == 0:
                A.pop(i)
                continue
            i += 1
            
    for a in A:
        P.append(a)
            
    return P
 
# 同じものを含む順列
def permutation_with_duplicates(L):
 
    if L == []:
        return [[]]
 
    else:
        ret = []
 
        # set（集合）型で重複を削除、ソート
        S = sorted(set(L))
 
        for i in S:
 
            data = L[:]
            data.remove(i)
 
            for j in permutation_with_duplicates(data):
                ret.append([i] + j)
 
        return ret
 
 
s= input()
f=s.index("A")
l=len(s)-1-s[::-1].index("Z")
print(l-f+1)

Measured memory_bytes: 5692