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

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

Input

The only line of the input contains three integers l, r and k (1 ≤ l ≤ r < 10^{18}, 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 to r such that each number contains at most k different digits, modulo 998244353.

Examples

Input

10 50 2


Output

1230


Input

1 2345 10


Output

2750685


Input

101 154 2


Output

2189

Note

For the first example the answer is just the sum of numbers from l to r which equals to (50 ⋅ 51)/(2) - (9 ⋅ 10)/(2) = 1230. This example also explained in the problem statement but for k = 1.

For the second example the answer is just the sum of numbers from l to r which equals to (2345 ⋅ 2346)/(2) = 2750685.

For the third example the answer is 101 + 110 + 111 + 112 + 113 + 114 + 115 + 116 + 117 + 118 + 119 + 121 + 122 + 131 + 133 + 141 + 144 + 151 = 2189.
/*
Keep solving problems.
*/

import java.util.*;
import java.io.*;

public class CFA {
    BufferedReader br;
    PrintWriter out;
    StringTokenizer st;
    boolean eof;
    private static final long MOD = 998244353L;
    private static final int[] dx = {0, -1, 0, 1};
    private static final int[] dy = {1, 0, -1, 0};
    private static final String yes = "Yes";
    private static final String no = "No";

    long l;
    long r;
    int upp;
    long[] pow10;
    int[] bit;
    void solve() throws IOException {
        l = nextLong();
        r = nextLong();
        upp = nextInt();
        pow10 = new long[50];
        pow10[0] = 1;
        for (int i = 1; i < pow10.length; i++) {
            pow10[i] = pow10[i - 1] * 10;
            pow10[i] %= MOD;
        }
        bit = new int[100000];
        for (int i = 0; i < bit.length; i++) {
            bit[i] = Integer.bitCount(i);
        }

        long res = add(calc(r), -calc(l - 1));
        outln(res);
    }

    long calc(long bound) {
        char[] arr = Long.toString(bound).toCharArray();
        int n = arr.length;
        int sz = 10;
        //i: the i first digit
        //j: is equal
        //k: has started
        //p: mask
        long[][][][] dp = new long[1 + n][2][2][1 << sz];
        long[][][][] sum = new long[1 + n][2][2][1 << sz];
        dp[0][1][0][0] = 1;

        for (int i = 0; i < n; i++) {
        int curDig = arr[i] - '0';
        for (int j = 0; j < 2; j++) {
        for (int k = 0; k < 2; k++) {
        for (int mask = 0; mask < (1 << sz); mask++) {
            if (dp[i][j][k][mask] == 0) {
                continue;
            }

            int high = j == 0 ? 9 : curDig;
            for (int nxt = 0; nxt <= high; nxt++) {
                int nj = 0;
                if (j == 1 && nxt == curDig) {
                    nj = 1;
                }
                int nk = 0;
                if (k == 1 || nxt != 0) {
                    nk = 1;
                }
                int nmask = mask;
                if (nk == 1) {
                    nmask = nmask | (1 << nxt);
                }

                dp[1 + i][nj][nk][nmask] = add(dp[1 + i][nj][nk][nmask], dp[i][j][k][mask]);
                long tmp = add(sum[i][j][k][mask], pow10[n - 1 - i] * nxt % MOD * dp[i][j][k][mask] % MOD);
                sum[1 + i][nj][nk][nmask] = add(sum[1 + i][nj][nk][nmask], tmp);
            }
        }
        }
        }
        }

        long res = 0;
        long cnt = 0;
        for (int j = 0; j < 2; j++) {
            for (int k = 0; k < (1 << sz); k++) {
                if (bit[k] <= upp) {
                    res = add(res, sum[n][j][1][k]);
                    cnt += dp[n][j][1][k];
                }
            }
        }

        return res;
    }

    long add(long v1, long v2) {
        v1 += v2;
        if (v1 >= MOD) {
            v1 -= MOD;
        }
        if (v1 < 0) {
            v1 += MOD;
        }

        return v1;
    }

    void shuffle(int[] a) {
        int n = a.length;
        for(int i = 0; i < n; i++) {
            int r = i + (int) (Math.random() * (n - i));
            int tmp = a[i];
            a[i] = a[r];
            a[r] = tmp;
        }
    }
    long gcd(long a, long b) {
        while(a != 0 && b != 0) {
            long c = b;
            b = a % b;
            a = c;
        }
        return a + b;
    }
    private void outln(Object o) {
        out.println(o);
    }
    private void out(Object o) {
        out.print(o);
    }
    private void formatPrint(double val) {
        System.out.format("%.9f%n", val);
    }
    public CFA() throws IOException {
        br = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        solve();
        out.close();
    }
    public static void main(String[] args) throws IOException {
        new CFA();
    }

    public long[] nextLongArr(int n) throws IOException{
        long[] res = new long[n];
        for(int i = 0; i < n; i++)
            res[i] = nextLong();
        return res;
    }
    public int[] nextIntArr(int n) throws IOException {
        int[] res = new int[n];
        for(int i = 0; i < n; i++)
            res[i] = nextInt();
        return res;
    }
    public String nextToken() {
        while (st == null || !st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                eof = true;
                return null;
            }
        }
        return st.nextToken();
    }
    public String nextString() {
        try {
            return br.readLine();
        } catch (IOException e) {
            eof = true;
            return null;
        }
    }
    public int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }
    public long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }
    public double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }
}