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.
#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
const long long N = 1e5;
long long l, r, k, d[20][1024][2][2], sum[20][1024][2][2];
long long dp(long long x) {
  long long cnt = 19;
  long long s[20];
  for (long long i = 0; i <= 19; i++) s[i] = 0;
  while (x) {
    s[cnt] = x % 10;
    x /= 10;
    cnt--;
  }
  for (long long z1 = 0; z1 < 20; z1++)
    for (long long z2 = 0; z2 < 1024; z2++) {
      d[z1][z2][0][0] = 0;
      d[z1][z2][1][0] = 0;
      d[z1][z2][0][1] = 0;
      d[z1][z2][1][1] = 0;
      sum[z1][z2][0][0] = 0;
      sum[z1][z2][1][0] = 0;
      sum[z1][z2][0][1] = 0;
      sum[z1][z2][1][1] = 0;
    }
  d[0][0][0][0] = 1;
  for (long long p = 1; p <= 19; p++) {
    for (long long m = 0; m < 1024; m++) {
      for (long long q = 0; q < 2; q++) {
        for (long long r = 0; r < 2; r++) {
          long long lim = 9;
          if (!q) lim = s[p];
          for (long long nx = 0; nx <= lim; nx++) {
            long long nm = m | (1 << nx);
            if (m == 0 && nm == 1) nm = 0;
            long long nq = q | (nx < s[p]);
            long long nr = r | (nx > 0);
            d[p][nm][nq][nr] += d[p - 1][m][q][r];
            d[p][nm][nq][nr] %= mod;
            sum[p][nm][nq][nr] += sum[p - 1][m][q][r] * 10;
            sum[p][nm][nq][nr] %= mod;
            sum[p][nm][nq][nr] += d[p - 1][m][q][r] * nx;
            sum[p][nm][nq][nr] %= mod;
          }
        }
      }
    }
  }
  long long res = 0;
  for (long long m = 0; m < 1024; m++) {
    if (__builtin_popcount(m) <= k) {
      res += sum[19][m][0][1];
      res %= mod;
      res += sum[19][m][1][1];
      res %= mod;
    }
  }
  return res;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> l >> r >> k;
  long long tmp = (dp(r) - dp(l - 1)) % mod;
  if (tmp < 0)
    cout << tmp + mod;
  else
    cout << tmp;
}
