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 int mo = 998244353;
int k, len, a[20], f[20][1024], g[20][1024], mi[20];
long long l, r;
pair<int, int> dfs(int t, int sta, bool top, bool zero) {
  if (t == 0) return make_pair(0, 1);
  if (!zero && !top && f[t][sta] > -1) return make_pair(f[t][sta], g[t][sta]);
  int lim = (top ? a[t] : 9), ans = 0, num = 0, one = 0;
  for (int i = 0; i < 10; i++)
    if (sta >> i & 1) one++;
  for (int i = 0; i <= lim; i++)
    if (one < k || (one == k && (sta >> i & 1)) ||
        (one == k && i == 0 && zero)) {
      pair<int, int> tmp = dfs(t - 1, (zero && i == 0 ? sta : (sta | (1 << i))),
                               top && i == lim, zero && i == 0);
      ans = ((long long)ans + tmp.first +
             (long long)tmp.second * i * mi[t - 1] % mo) %
            mo;
      num = (num + tmp.second) % mo;
    }
  if (!zero && !top) {
    f[t][sta] = ans;
    g[t][sta] = num;
  }
  return make_pair(ans, num);
}
int work(long long x) {
  if (!x) return 0;
  len = 0;
  while (x) {
    a[++len] = x % 10;
    x /= 10;
  }
  return dfs(len, 0, true, true).first;
}
int main() {
  scanf("%lld%lld%d", &l, &r, &k);
  memset(f, -1, sizeof(f));
  mi[0] = 1;
  for (int i = 1; i <= 19; i++) mi[i] = (long long)mi[i - 1] * 10 % mo;
  printf("%d\n", (work(r) - work(l - 1) + mo) % mo);
  return 0;
}
