CodeForces

You are given two integers l l and r r (l≤r l≤r ). Your task is to calculate the sum of numbers from l l to r r (including l l and r r ) such that each number contains at most k k different digits, and print this sum modulo 998244353 998244353 .

For example, if k=1 k=1 then you have to calculate all numbers from l l to r r such that each number is formed using only one digit. For l=10,r=50 l=10,r=50 the answer is 11+22+33+44=110 11+22+33+44=110 .

Input

The only line of the input contains three integers l l , r r and k k (1≤l≤r<10 18 ,1≤k≤10 1≤l≤r<1018,1≤k≤10 ) — the borders of the segment and the maximum number of different digits.

Output

Print one integer — the sum of numbers from l l to r r such that each number contains at most k k different digits, modulo 998244353 998244353 .

Examples

Input

10 50 2

Output

1230

Input

1 2345 10

Output

2750685

Input

101 154 2

Output

2189

题意：求区间[L,R]的满足digit种类不超过K的数字之和。

思路：与常规我数位DP不一样的是，这里求是不是个数，而是这些书之和。所以我们要记录一个二元组(x,y)分别表示（子树之和，子树叶子个数）。

（数位DP其实就是一棵树，子树相同的时候可以直接调用答案）

那么当前节点为根的树的信息=（所有子树的和+当前位*叶子个数，所有叶子个数之和）。

#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define ll long long
#define pii pair<int,int>
#define mp make_pair
using namespace std;
const int Mod=998244353;
pii dp[20][1<<10],O=mp(0,0);
int v[20],num[1<<10],K,q[20],tot;
pii dfs(int pos,int st,int lim)
{
    if(!lim&&dp[pos][st]!=O) return dp[pos][st];
    if(pos==1) return mp(0,1);
    int up=9; pii res=O,tmp; if(lim) up=q[pos-1];
    rep(i,0,up){
        if(num[st|(1<<i)]<=K){
            tmp=dfs(pos-1,st|(1<<i),lim&&i==up);
            (res.second+=tmp.second)%=Mod; 
            (res.first+=tmp.first)%=Mod;
            (res.first+=(ll)v[pos-1]*i%Mod*tmp.second%Mod)%=Mod;
        }
    }
    return dp[pos][st]=res;
}
int cal(ll x)
{
    if(x==0) return 0;
    tot=0; int ans=0;
    while(x) q[++tot]=x%10,x/=10;
    memset(dp,0,sizeof(dp));
    rep(i,1,tot){
        ll up=9; if(i==tot) up=q[tot];
        rep(j,1,up){
            pii tmp=dfs(i,1<<j,(i==tot)&&(j==q[tot]));
            (ans+=(ll)v[i]*j%Mod*tmp.second%Mod)%=Mod;
            (ans+=tmp.first)%=Mod;
        }
    }
    return ans;
}
int main()
{
    rep(i,1,1<<10) num[i]=num[i>>1]+(i&1);
    v[1]=1; rep(i,2,18) v[i]=(ll)v[i-1]*10%Mod;
    ll L,R; scanf("%lld%lld%d",&L,&R,&K);
    printf("%d
",((cal(R)-cal(L-1))%Mod+Mod)%Mod);
    return 0;
}