import os,sys,random,threading
#sys.exit() 退出程序
#sys.setrecursionlimit(10**6) #调整栈空间
from random import randint,choice,shuffle
#randint(a,b)从[a,b]范围随机选择一个数
#choice(seq)seq可以是一个列表,元组或字符串,从seq中随机选取一个元素
#shuffle(x)将一个可变的序列x中的元素打乱
from copy import deepcopy
from io import BytesIO,IOBase
from types import GeneratorType
from functools import lru_cache,reduce
#reduce(op,迭代对象)
from bisect import bisect_left,bisect_right
#bisect_left(x) 大于等于x的第一个下标
#bisect_right(x) 大于x的第一个下标
from collections import Counter,defaultdict,deque
from itertools import accumulate,combinations,permutations
#accumulate(a)用a序列生成一个累积迭代器，一般list化前面放个[0]做前缀和用
#combinations(a,k)a序列选k个 组合迭代器
#permutations(a,k)a序列选k个 排列迭代器
from heapq import  heapify,heappop,heappush
#heapify将列表转为堆
from typing import Generic,Iterable,Iterator,TypeVar,Union,List
from string import ascii_lowercase,ascii_uppercase,digits
#小写字母，大写字母，十进制数字
from math import ceil,floor,sqrt,pi,factorial,gcd,log,log10,log2,inf
#ceil向上取整，floor向下取整 ，sqrt开方 ，factorial阶乘
from decimal import Decimal,getcontext
#Decimal(s) 实例化Decimal对象,一般使用字符串
#getcontext().prec=100 修改精度
from sys import stdin, stdout, setrecursionlimit
input = lambda: sys.stdin.readline().rstrip("\r\n")
MI = lambda :map(int,input().split())
li = lambda :list(MI())
ii = lambda :int(input())
mod = int(1e9 + 7) #998244353
inf = 1<<60
py = lambda :print("YES")
pn = lambda :print("NO")
DIRS = [(0, 1), (1, 0), (0, -1), (-1, 0)]  # 右下左上
DIRS8 = [(0, 1), (1, 1), (1, 0), (1, -1), (0, -1), (-1, -1), (-1, 0),(-1, 1)]  # →↘↓↙←↖↑↗

class PrimeTable:
    def __init__(self, n=int((10**9)**0.5)+3) -> None:
        #值域1e9的话考虑质因子，只需小于等于(10**9)**0.5的质数即可
        #任意一个正整数n最多只有一个质因子大于根号n
        self.n = n
        self.primes = [] #小于等于n的所有质数
        self.min_div = [0] * (n+1)
        self.min_div[1] = 1
 
        mu = [0] * (n+1)
        phi = [0] * (n+1)
        mu[1] = 1
        phi[1] = 1
 
        for i in range(2, n+1):
            if not self.min_div[i]:
                self.primes.append(i)
                self.min_div[i] = i
                mu[i] = -1
                phi[i] = i-1
            for p in self.primes:
                if i * p > n: break
                self.min_div[i*p] = p
                if i % p == 0:
                    phi[i*p] = phi[i] * p
                    break
                else:
                    mu[i*p] = -mu[i]
                    phi[i*p] = phi[i] * (p - 1)
    # x是否质数
    def is_prime(self, x:int):
        if x < 2: return False
        if x <= self.n: return self.min_div[x] == x
        for p in self.primes:
            if p * p > x: break
            if x % p == 0: return False
        return True

    # x分解质因数:[p, cnt] 质因子p，个数cnt
    # 用的yield，当作一个可遍历的数据对象
    #一个数一定可以分解为多个质数的连乘积
    #n = x^a * y^b * z^c ...  （x,y,z为质因数） n的约数个数=(a+1)(b+1)...(y+1)

    def prime_factorization(self, x:int):
        for p in self.primes:
            if p * p > x: break
            if x <= self.n: break
            if x % p == 0:
                cnt = 0
                while x % p == 0: cnt += 1; x //= p
                yield p, cnt
        while (1 < x and x <= self.n):
            p, cnt = self.min_div[x], 0
            while x % p == 0: cnt += 1; x //= p
            yield p, cnt
        if x >= self.n and x > 1:
            #小于等于(10**9)**0.5的质数除干净了，如果还大于1
            # 那么余下的数一定是一个大于等于n的质数
            yield x, 1
            
    # x的所有因数
    def get_factors(self, x:int):
        factors = [1]
        for p, b in self.prime_factorization(x):
            n = len(factors)
            for j in range(1, b+1):
                for d in factors[:n]:
                    factors.append(d * (p ** j))
        return factors


def getMu(n):
    mu = [0] * (n + 1)
    flg = [0] * (n + 1)
    p = [0] * (n + 1)
    tot = 0
    mu[1] = 1
    for i in range(2, n + 1):
        if flg[i] == 0:
            tot = tot + 1; p[tot] = i; mu[i] = -1
        j = 1
        while j <= tot and i * p[j] <= n:
            flg[i * p[j]] = 1
            if i % p[j] == 0:
                mu[i * p[j]] = 0
                break
            mu[i * p[j]] =  - mu[i]
            j = j + 1
    return mu

# f[n]表示恰好使用n个有标号的元素形成特定结构的方案数
# g[n]从n个中选出i个选出特定结构的总方案数


# 情况1：g[n]表示至多为n个
# g[n]=ΣC(n,i) * f[i] i∈[0,n]
# 反演
# f[n]=Σ(-1)^(n-i) * C(n,i) * g[i]   i∈[0,n]

# 情况2：g[k]表示至少为k个
# g[k]=ΣC(i,k) * f[i] i∈[k,n]
# 反演
# f[k]=Σ(-1)^(i-k) * C(i,k) * g[i]   i∈[k,n]

mod=998244353

pt=PrimeTable(10**6+1)

mb=getMu(10**6+1)

n=ii()

arr=li()

ans=0

cnt=[0]*(10**6+1)


tot=0

for a in arr:
    val=[1]
    for p,_ in pt.prime_factorization(a):
        for i in range(len(val)):
            val.append(val[i]*p)
    dp=0
    for v in val:
        dp+=cnt[v]*mb[v]
        dp%=mod
    #print(cnt[:10],val,mb[:10])
    #print(a,dp,tot)
    dp=(tot-dp+1)%mod
    tot=(tot+dp)%mod
    for v in val:
        cnt[v]+=dp
        cnt[v]%=mod
    #print(a,dp,tot)
print(tot)