//题目回忆：给定一串序列，要求两两合并，求最后结果%MOD
//不好算时，一定考虑贡献法，每一个数对答案的共享为，(a1+a2..+a[i-1])*10^g(a[i])+（i-1)*a[i];
//因此前缀和优化即可
//注意每一步都要取模，不然数太大，每一步都会爆
#include <iostream>
#include <algorithm>
#include <cmath>

using namespace std;

const int N = 3e5+10,MOD = 998244353;

typedef long long LL;

LL n;
LL a[N];
int c[N];
LL s[N];
LL sum;

int cale(int x)
{
    int res=0;
    while(x)
    {
        x/=10;
        res++;
    }
    return res;
}

LL p(int x)
{
    LL res=1;
    while(x)
    {
        res=res*10%MOD;
        x--;
    }
    return res;
}

int main()
{
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
        s[i]=(s[i-1]+a[i])%MOD;
        c[i]=cale(a[i]);
    }
    LL sum=0;
    for(int i=2;i<=n;i++)
        sum=(sum+(s[i-1]*(LL)p(c[i]))%MOD+((LL)(i-1)*a[i])%MOD)%MOD;
    cout<<sum;
}