#include <bits/stdc++.h>
using namespace std;
const long long MOD = 998244353;
long long dp[23][2][1 << 10][2];
long long calc(string s, long long k) {
  memset(dp, 0, sizeof(dp));
  dp[0][1][0][0] = 1;
  for (int pref = 0; pref < s.size(); ++pref) {
    for (int flag = 0; flag < 2; ++flag) {
      for (int mask = 0; mask < (1 << 10); ++mask) {
        for (int digit = 0; digit < 10; ++digit) {
          if (flag && digit > (s[pref] - '0')) {
            continue;
          }
          long long new_mask = (mask == 1 ? 0 : mask) | (1 << digit);
          long long new_flag = flag & (digit == (s[pref] - '0'));
          dp[pref + 1][new_flag][new_mask][0] += dp[pref][flag][mask][0];
          dp[pref + 1][new_flag][new_mask][1] +=
              (10LL * dp[pref][flag][mask][1] +
               1LL * digit * dp[pref][flag][mask][0]) %
              MOD;
          dp[pref + 1][new_flag][new_mask][0] %= MOD;
          dp[pref + 1][new_flag][new_mask][1] %= MOD;
        }
      }
    }
  }
  long long ans = 0;
  for (int flag = 0; flag < 2; ++flag) {
    for (int mask = 0; mask < (1 << 10); ++mask) {
      if (__builtin_popcount(mask) <= k) {
        ans = (ans + dp[s.size()][flag][mask][1]) % MOD;
      }
    }
  }
  return ans;
}
int main() {
  long long l, r, k;
  cin >> l >> r >> k;
  cout << (calc(to_string(r), k) - calc(to_string(l - 1), k) + MOD) % MOD
       << '\n';
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 dp[23][2][1 << 10][2];
long long calc(string s, long long k) {
  memset(dp, 0, sizeof(dp));
  dp[0][1][0][0] = 1;
  for (int pref = 0; pref < s.size(); ++pref) {
    for (int flag = 0; flag < 2; ++flag) {
      for (int mask = 0; mask < (1 << 10); ++mask) {
        for (int digit = 0; digit < 10; ++digit) {
          if (flag && digit > (s[pref] - '0')) {
            continue;
          }
          long long new_mask = (mask == 1 ? 0 : mask) | (1 << digit);
          long long new_flag = flag & (digit == (s[pref] - '0'));
          dp[pref + 1][new_flag][new_mask][0] += dp[pref][flag][mask][0];
          dp[pref + 1][new_flag][new_mask][1] +=
              (10LL * dp[pref][flag][mask][1] +
               1LL * digit * dp[pref][flag][mask][0]) %
              MOD;
          dp[pref + 1][new_flag][new_mask][0] %= MOD;
          dp[pref + 1][new_flag][new_mask][1] %= MOD;
        }
      }
    }
  }
  long long ans = 0;
  for (int flag = 0; flag < 2; ++flag) {
    for (int mask = 0; mask < (1 << 10); ++mask) {
      if (__builtin_popcount(mask) <= k) {
        ans = (ans + dp[s.size()][flag][mask][1]) % MOD;
      }
    }
  }
  return ans;
}
int main() {
  long long l, r, k;
  cin >> l >> r >> k;
  cout << (calc(to_string(r), k) - calc(to_string(l - 1), k) + MOD) % MOD
       << '\n';
  return 0;
}

```