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;
long long L, R;
int k;
int f[27][2][1027], g[27][2][1027], s[27], bit[1027];
long long num[27];
void pre() {
  memset(f, 0, sizeof(f));
  memset(g, 0, sizeof(g));
  memset(s, 0, sizeof(s));
  memset(bit, 0, sizeof(bit));
  bit[1] = 1;
  for (int i = 2; i < 1024; i++) bit[i] = bit[i / 2] + (i % 2 == 1);
  s[0] = 1;
  for (int i = 1; i <= 20; i++) s[i] = (10LL * s[i - 1]) % 998244353;
  g[0][0][0] = 1;
  g[0][1][0] = 1;
  for (int i = 1; i <= 20; i++)
    for (int j = 0; j <= 9; j++) {
      for (int k = 0; k < 1024; k++) {
        g[i][1][k | (1 << j)] =
            (g[i][1][k | (1 << j)] + g[i - 1][1][k]) % 998244353;
        if (j == 0)
          g[i][0][k] = (g[i][0][k] + g[i - 1][0][k]) % 998244353;
        else
          g[i][0][k | (1 << j)] =
              (g[i][0][k | (1 << j)] + g[i - 1][1][k]) % 998244353;
        f[i][1][k | (1 << j)] =
            (f[i][1][k | (1 << j)] + f[i - 1][1][k]) % 998244353;
        f[i][1][k | (1 << j)] =
            (f[i][1][k | (1 << j)] +
             (1LL * j * (1LL * s[i - 1] * g[i - 1][1][k]) % 998244353) %
                 998244353) %
            998244353;
        if (j == 0)
          f[i][0][k] = (f[i][0][k] + f[i - 1][0][k]) % 998244353;
        else {
          f[i][0][k | (1 << j)] =
              (f[i][0][k | (1 << j)] + f[i - 1][1][k]) % 998244353;
          f[i][0][k | (1 << j)] =
              (f[i][0][k | (1 << j)] +
               (1LL * j * (1LL * s[i - 1] * g[i - 1][1][k]) % 998244353) %
                   998244353) %
              998244353;
        }
      }
    }
}
int calc(int now, long long val, int pos) {
  int res = 0;
  if (now == 0 || now == 1) {
    for (int i = 0; i < 1024; i++)
      if (bit[i] <= k) res = (res + f[pos][0][i]) % 998244353;
  } else {
    for (int i = 0; i < 1024; i++)
      if (bit[i | now] <= k) {
        res = (res + f[pos][1][i]) % 998244353;
        res = (res + (1LL * (val % 998244353) * g[pos][1][i]) % 998244353) %
              998244353;
      }
  }
  return (res + 998244353) % 998244353;
}
int solve(long long x, int k) {
  int ans = 0, pos = 0;
  memset(num, 0, sizeof(num));
  long long tmp = x;
  while (tmp > 0) {
    num[++pos] = tmp % 10;
    tmp /= 10;
  }
  int now = 0;
  long long val = 0LL;
  for (int i = 1; i < 20 - i; i++) swap(num[i], num[20 - i]);
  for (int i = 1; i <= 19; i++) {
    for (int j = 0; j < num[i]; j++)
      ans = (ans + calc(now | (1 << j), val + 1LL * j * s[19 - i], 19 - i)) %
            998244353;
    val += 1LL * num[i] * s[19 - i];
    if (val != 0LL) now |= (1 << num[i]);
  }
  return ans;
}
int main() {
  pre();
  scanf("%lld%lld%d", &L, &R, &k);
  printf("%d\n", (solve(R + 1, k) + 998244353 - solve(L, k)) % 998244353);
  return 0;
}
