#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
long long cnt[20][2][2][1 << 11], sm[20][2][2][1 << 11], l, r, k, pw[20] = {1};
string s;
inline long long cntbits(long long wtf) {
  long long res = 0;
  while (wtf) {
    res++;
    wtf -= (wtf & -wtf);
  }
  return res;
}
long long solve(long long d, long long z, long long m, long long msk) {
  if (d == s.size())
    return cntbits(msk) <= k;
  else if (cnt[d][z][m][msk] != -1)
    return cnt[d][z][m][msk];
  cnt[d][z][m][msk] = 0;
  for (int i = 0; i <= (m ? s[d] - '0' : 9); i++) {
    if (i == 0 && z) {
      cnt[d][z][m][msk] = (cnt[d][z][m][msk] + solve(d + 1, 1, 0, msk)) % mod;
      sm[d][z][m][msk] = (sm[d][z][m][msk] + sm[d + 1][1][0][msk]) % mod;
    } else {
      cnt[d][z][m][msk] =
          (cnt[d][z][m][msk] +
           solve(d + 1, 0, i == s[d] - '0' && m, msk | (1 << i))) %
          mod;
      sm[d][z][m][msk] =
          (sm[d][z][m][msk] +
           solve(d + 1, 0, i == s[d] - '0' && m, msk | (1 << i)) * i % mod *
               pw[s.size() - d - 1] % mod) %
          mod;
      sm[d][z][m][msk] = (sm[d][z][m][msk] +
                          sm[d + 1][0][i == s[d] - '0' && m][msk | (1 << i)]) %
                         mod;
    }
  }
  return cnt[d][z][m][msk];
}
int main() {
  scanf("%lld%lld%lld", &l, &r, &k);
  s = to_string(r);
  for (int i = 1; i < 20; i++) pw[i] = (pw[i - 1] * 10) % mod;
  memset(cnt, -1, sizeof(cnt));
  solve(0, 1, 1, 0);
  long long ans = sm[0][1][1][0];
  s = to_string(l - 1);
  memset(cnt, -1, sizeof(cnt));
  memset(sm, 0, sizeof(sm));
  solve(0, 1, 1, 0);
  ans -= sm[0][1][1][0];
  if (ans < 0) ans += mod;
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
long long cnt[20][2][2][1 << 11], sm[20][2][2][1 << 11], l, r, k, pw[20] = {1};
string s;
inline long long cntbits(long long wtf) {
  long long res = 0;
  while (wtf) {
    res++;
    wtf -= (wtf & -wtf);
  }
  return res;
}
long long solve(long long d, long long z, long long m, long long msk) {
  if (d == s.size())
    return cntbits(msk) <= k;
  else if (cnt[d][z][m][msk] != -1)
    return cnt[d][z][m][msk];
  cnt[d][z][m][msk] = 0;
  for (int i = 0; i <= (m ? s[d] - '0' : 9); i++) {
    if (i == 0 && z) {
      cnt[d][z][m][msk] = (cnt[d][z][m][msk] + solve(d + 1, 1, 0, msk)) % mod;
      sm[d][z][m][msk] = (sm[d][z][m][msk] + sm[d + 1][1][0][msk]) % mod;
    } else {
      cnt[d][z][m][msk] =
          (cnt[d][z][m][msk] +
           solve(d + 1, 0, i == s[d] - '0' && m, msk | (1 << i))) %
          mod;
      sm[d][z][m][msk] =
          (sm[d][z][m][msk] +
           solve(d + 1, 0, i == s[d] - '0' && m, msk | (1 << i)) * i % mod *
               pw[s.size() - d - 1] % mod) %
          mod;
      sm[d][z][m][msk] = (sm[d][z][m][msk] +
                          sm[d + 1][0][i == s[d] - '0' && m][msk | (1 << i)]) %
                         mod;
    }
  }
  return cnt[d][z][m][msk];
}
int main() {
  scanf("%lld%lld%lld", &l, &r, &k);
  s = to_string(r);
  for (int i = 1; i < 20; i++) pw[i] = (pw[i - 1] * 10) % mod;
  memset(cnt, -1, sizeof(cnt));
  solve(0, 1, 1, 0);
  long long ans = sm[0][1][1][0];
  s = to_string(l - 1);
  memset(cnt, -1, sizeof(cnt));
  memset(sm, 0, sizeof(sm));
  solve(0, 1, 1, 0);
  ans -= sm[0][1][1][0];
  if (ans < 0) ans += mod;
  printf("%lld\n", ans);
  return 0;
}

```