#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<stack>
#include<map>
#include<vector>
#include<queue>
#include<set>
#include<iomanip>
#include<cctype>
#include<ctime>
using namespace std;
typedef long long ll;
#define edl putchar('\n')
#define sscc ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define ROF(i,a,b) for(int i=a;i>=b;i--)
#define FORLL(i,a,b) for(ll i=a;i<=b;i++)
#define ROFLL(i,a,b) for(ll i=a;i>=b;i--)
#define mst(a) memset(a,0,sizeof(a))
#define mstn(a,n) memset(a,n,sizeof(a))
#define zero(x)(((x)>0?(x):-(x))<eps)
#define si(a) scanf("%d",&a)
#define sl(a) scanf("%lld",&a)
#define sd(a) scanf("%lf",&a)
#define ss(a) scanf("%s",a)
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
int month[15]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int dir[8][2]={{-1,0},{1,0},{0,-1},{0,1},{-1,-1},{1,1},{1,-1},{-1,1}};
const int MAXN=2e5+5;
const int INF=1<<30;
const long long mod=998244353;
const double eps=1e-8;
const ll inff=0x3f3f3f3f3f3f3f3f;
ll dp[25][1030][12];/*记录答案总和的数组*/
ll d[25][1030];/*记录答案个数的数组*/
ll t[25];/*10的幂次方，节省时间*/
int a[25];/*记录每一位数字*/
int f[25];/*2的幂次方，节省时间*/
int k[1030];/*状态压缩，记录某个状态有多少个不同的数字*/
int p;/*题目问的最大不同数量*/
struct num
{ll a,b;/*a对应d数组，b对应dp数组*/
}s;
num dfs(int pos,int now,int sta,int limit)
{if(pos==0){return (num){1,now};} if(!limit&&dp[pos][sta][now]!=-1){return (num){d[pos][sta],dp[pos][sta][now]};}int up=limit?a[pos]:9,g;ll ans=0,cnt=0;FOR(i,0,up){if(sta==0&&i==0)/*前导0的特殊判断*/ {s=dfs(pos-1,i,sta,limit&&i==up);cnt+=s.a;/*答案个数*/ans+=s.b;/*答案总和*/cnt%=mod;ans%=mod;}else{g=sta|f[i];if(k[g]<=p){s=dfs(pos-1,i,g,limit&&i==up);cnt+=s.a;/*答案个数*/ans+=s.b;/*答案总和*/cnt%=mod;ans%=mod;}}    }ans=(ans+cnt*t[pos]%mod*(ll)now%mod)%mod;if(!limit)d[pos][sta]=cnt,dp[pos][sta][now]=ans;return (num){cnt,ans};
}
ll solve(ll x)
{if(x==0) return 0;int pos=0;while(x){a[++pos]=x%10;x/=10;}a[pos+1]=0;return dfs(pos,0,0,1).b;
}
int main()
{ll n,m;f[0]=1,t[0]=1;FOR(i,1,18)f[i]=f[i-1]*2,t[i]=t[i-1]*10,t[i]%=mod;FOR(i,0,1024){int j=i,cnt=0;while(j){cnt+=j%2;j/=2;}k[i]=cnt;}FOR(i,0,20)FOR(j,0,1024){d[i][j]=-1;FOR(I,0,9)dp[i][j][I]=-1;}//scanf("%d",&T);//while(1){scanf("%lld%lld%d",&n,&m,&p);n=solve(n-1);m=solve(m);printf("%lld\n",(m-n+mod)%mod);}return 0;
}