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;
long long r;
long long s[100];
pair<long long, long long> dp[100][1 << 10];
long long len;
long long T;
long long fac[20];
long long cnt[1 << 10];
long long k;
pair<long long, long long> dfs(long long pos, int msk, int lead, int limit) {
  if (pos < 1) return make_pair(1LL, 0LL);
  if ((dp[pos][msk].first != -1 && (!limit) && (!lead))) return dp[pos][msk];
  pair<long long, long long> ret = make_pair(0LL, 0LL);
  long long res = limit ? s[pos] : 9;
  for (long long i = 0; i <= res; i++) {
    if ((!i) && lead) {
      pair<long long, long long> tmp = dfs(pos - 1, 0, 1, limit && (i == res));
      (ret.first += tmp.first) %= 998244353;
      (ret.second += tmp.second) %= 998244353;
    } else {
      int nxt = msk | (1 << i);
      if (cnt[nxt] > k) continue;
      pair<long long, long long> tmp =
          dfs(pos - 1, nxt, 0, limit && (i == res));
      (ret.first += tmp.first) %= 998244353;
      (ret.second += tmp.second + i * fac[pos - 1] % 998244353 * tmp.first %
                                      998244353) %= 998244353;
    }
  }
  if (!limit && !lead) dp[pos][msk] = ret;
  return ret;
}
pair<long long, long long> part(long long x) {
  len = 0;
  while (x) s[++len] = x % 10, x /= 10;
  return dfs(len, 0, 1, 1);
}
int main() {
  fac[0] = 1LL;
  for (long long i = 1; i < 20; ++i) fac[i] = fac[i - 1] * 10LL % 998244353;
  for (int i = 0; i < (1 << 10); ++i) {
    int cur = 0;
    for (int j = 0; j < 10; ++j) {
      if (i & (1 << j)) ++cur;
    }
    cnt[i] = 1LL * cur;
  }
  memset(dp, -1, sizeof dp);
  scanf("%lld %lld", &l, &r);
  scanf("%lld", &k);
  long long ans = part(r).second - part(l - 1).second;
  ans = (ans + 998244353) % 998244353;
  printf("%lld\n", ans);
  return 0;
}
