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, k, num[23], pov[23], tp;
pair<long long, long long> f[23][1 << 10];
pair<long long, long long> dfs(long long pos, long long sta, bool lim,
                               bool lead) {
  if (__builtin_popcount(sta) > k) return make_pair(0, 0);
  if (!pos) return make_pair(0, 1);
  if (!lim && !lead && f[pos][sta] != make_pair(-1ll, -1ll)) return f[pos][sta];
  pair<long long, long long> res = make_pair(0, 0);
  for (long long i = 0; i <= (lim ? num[pos] : 9); i++) {
    pair<long long, long long> tmp =
        dfs(pos - 1, sta | (lead && !i ? 0 : (1 << i)), lim && (i == num[pos]),
            lead && (!i));
    (res.first += tmp.first) %= mod;
    (res.second += tmp.second) %= mod;
    (res.first += tmp.second * i % mod * pov[pos] % mod) %= mod;
  }
  if (!lim && !lead) f[pos][sta] = res;
  return res;
}
long long calc(long long ip) {
  tp = 0;
  while (ip) num[++tp] = ip % 10, ip /= 10;
  return dfs(tp, 0, 1, 1).first;
}
signed main() {
  scanf("%lld%lld%lld", &l, &r, &k), l--, memset(f, -1, sizeof(f));
  pov[1] = 1;
  for (long long i = 2; i < 23; i++) pov[i] = (pov[i - 1] * 10) % mod;
  printf("%lld\n", (calc(r) - calc(l) + mod) % mod);
  return 0;
}
