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>
const long long P(998244353);
inline void inc(long long &x, long long y) {
  if ((x += y) >= P) x -= P;
}
long long k, a[22], pw10[22];
long long f[22][1033][2][2], g[22][1033][2][2];
long long dfs(long long p, long long s, bool lim, bool lead) {
  long long &ans = g[p][s][lim][lead];
  if (~ans) return ans;
  if (__builtin_popcount(s) > k) return f[p][s][lim][lead] = ans = 0;
  if (!p) return f[p][s][lim][lead] = 1, ans = 0;
  f[p][s][lim][lead] = ans = 0;
  long long up = lim ? a[p] : 9;
  for (long long i(0); i <= (up); ++i) {
    long long to = (s >> i & 1 ^ 1) && (i || !lead) ? s | 1 << i : s;
    long long v = dfs(p - 1, to, lim && i == up, lead && i == 0);
    ans = (ans + v +
           1ll * i * pw10[p - 1] % P *
               f[p - 1][to][lim && i == up][lead && i == 0]) %
          P;
    inc(f[p][s][lim][lead], f[p - 1][to][lim && i == up][lead && i == 0]);
  }
  return ans;
}
long long solve(long long x) {
  if (!x) return 0;
  for (a[0] = 0; x; x /= 10) a[++a[0]] = x % 10;
  memset(g, -1, sizeof g);
  return dfs(a[0], 0, 1, 1);
}
signed main() {
  pw10[0] = 1;
  for (long long i(1); i <= (19); ++i) pw10[i] = 10ll * pw10[i - 1] % P;
  long long l, r;
  std::cin >> l >> r >> k;
  std::cout << (solve(r) - solve(l - 1) + P) % P;
  return 0;
}
