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;
long long l, r;
long long k, f[20][1 << 10], g[20][1 << 10], a[20], cnt, P[20];
long long read() {
  long long x = 0, f = 1;
  char s;
  while ((s = getchar()) > '9' || s < '0')
    if (s == '-') f = -1;
  while (s >= '0' && s <= '9')
    x = (x << 1) + (x << 3) + (s ^ 48), s = getchar();
  return x * f;
}
pair<long long, long long> DP(const long long pos, const long long sta,
                              const bool zero, const bool lim) {
  if (!pos) return make_pair(__builtin_popcount(sta) <= k, 0);
  if (!zero && (!lim) && (~f[pos][sta]))
    return make_pair(f[pos][sta], g[pos][sta]);
  long long up = lim ? a[pos] : 9;
  pair<long long, long long> ans = make_pair(0, 0);
  for (long long i = 0; i <= up; ++i) {
    long long Sta = (zero && i == 0) ? sta : (sta | (1 << i));
    pair<long long, long long> ret =
        DP(pos - 1, Sta, zero && i == 0, lim && i == up);
    ans = make_pair(
        (ans.first + ret.first) % mod,
        ((1ll * P[pos - 1] * i % mod * ret.first % mod + ret.second) % mod +
         ans.second) %
            mod);
  }
  if (!zero && (!lim)) f[pos][sta] = ans.first, g[pos][sta] = ans.second;
  return ans;
}
long long solve(long long x) {
  cnt = 0;
  memset(f, -1, sizeof f);
  memset(g, -1, sizeof g);
  while (x) a[++cnt] = x % 10, x /= 10;
  return DP(cnt, 0, 1, 1).second;
}
signed main() {
  l = read(), r = read(), k = read();
  P[0] = 1;
  for (long long i = 1; i <= 18; ++i) P[i] = 10ll * P[i - 1] % mod;
  printf("%lld\n", (solve(r) - solve(l - 1) + mod) % mod);
  return 0;
}
