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.
// upsolve with rainboy
import java.io.*;
import java.util.*;

public class CF1073E {
	static final int MD = 998244353, N = 18;
	static int[][] cnt = new int[1 + N][1 << 10];
	static int[][] sum = new int[1 + N][1 << 10];
	static long[] pow10 = new long[1 + N];
	static void init() {
		long p = 1;
		for (int n = 0; n <= N; n++) {
			pow10[n] = p;
			p *= 10;
		}
		cnt[0][0] = 1;
		for (int n = 1; n <= N; n++) {
			int[] c1 = cnt[n - 1];
			int[] s1 = sum[n - 1];
			p = pow10[n - 1] % MD;
			for (int b = 1; b < 1 << 10; b++)
				for (int d = 0; d < 10; d++)
					if ((b & 1 << d) != 0) {
						int c = (c1[b] + c1[b ^ 1 << d]) % MD;
						int s = (s1[b] + s1[b ^ 1 << d]) % MD;
						s = (int) ((p * d % MD * c + s) % MD);
						cnt[n][b] = (cnt[n][b] + c) % MD;
						sum[n][b] = (sum[n][b] + s) % MD;
					}
		}
	}
	static int dcount(long x) {
		return x == 0 ? 0 : 1 + dcount(x / 10);
	}
	static void solve(long x, int n, int[] cc, int[] ss, int one) {
		if (n == 0)
			return;
		long p = pow10[n - 1];
		int[] cc_ = new int[1 << 10];
		int[] ss_ = new int[1 << 10];
		solve(x % p, n - 1, cc_, ss_, 0);
		int a = (int) (x / p);
		p %= MD;
		for (int b = 0; b < 1 << 10; b++)
			for (int d = one; d <= a; d++)
				if ((b & 1 << d) != 0) {
					int[] c1, s1;
					if (d == a) {
						c1 = cc_;
						s1 = ss_;
					} else {
						c1 = cnt[n - 1];
						s1 = sum[n - 1];
					}
					int c = (c1[b] + c1[b ^ 1 << d]) % MD;
					int s = (s1[b] + s1[b ^ 1 << d]) % MD;
					s = (int) ((p * d % MD * c + s) % MD);
					cc[b] = (cc[b] + c) % MD;
					ss[b] = (ss[b] + s) % MD;
				}
	}
	static int solve_(long x, int k) {
		int n = dcount(x);
		int[] cc_ = new int[1 << 10];
		int[] ss_ = new int[1 << 10];
		solve(x, n, cc_, ss_, 1);
		int ans = 0;
		for (int b = 1; b < 1 << 10; b++) {
			int k_ = 0;
			for (int d = 0; d < 10; d++)
				if ((b & 1 << d) != 0)
					k_++;
			if (k_ <= k) {
				ans = (ans + ss_[b]) % MD;
				for (int i = 0; i < n - 1; i++) {
					int[] c1 = cnt[i];
					int[] s1 = sum[i];
					long p = pow10[i] % MD;
					for (int d = 1; d < 10; d++)
						if ((b & 1 << d) != 0) {
							int c = (c1[b] + c1[b ^ 1 << d]) % MD;
							int s = (s1[b] + s1[b ^ 1 << d]) % MD;
							s = (int) ((p * d % MD * c + s) % MD);
							ans = (ans + s) % MD;
						}
				}
			}
		}
		return ans;
	}
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(br.readLine());
		long l = Long.parseLong(st.nextToken());
		long r = Long.parseLong(st.nextToken());
		int k = Integer.parseInt(st.nextToken());
		init();
		System.out.println((solve_(r + 1, k) - solve_(l, k) + MD) % MD);
	}
}
