#include <bits/stdc++.h>
using namespace std;
long long l, r, k;
long long dp1[21][2][1 << 12], dp2[21][2][1 << 12];
std::vector<long long> num;
long long pow1[22];
void get(long long x) {
  while (x) {
    num.push_back(x % 10);
    x /= 10;
  }
  reverse((num).begin(), (num).end());
}
long long func(long long pos, long long flag, long long mask) {
  if (__builtin_popcount(mask) > k) return 0;
  if (pos == num.size()) {
    return 1;
  }
  if (dp1[pos][flag][mask] != -1) return dp1[pos][flag][mask];
  long long res = 0, lm = -1;
  if (!flag) {
    lm = num[pos];
  } else {
    lm = 9;
  }
  long long z;
  if (pos == 0)
    z = 1;
  else
    z = 0;
  long long ans = 0;
  for (long long int i = 0; i < lm + 1; i++) {
    long long nflag = flag;
    if (i < lm && flag == 0) nflag = 1;
    long long nmask = (mask | (1 << i));
    if (mask == 0 && i == 0) nmask = mask;
    ans += (((i * pow1[num.size() - pos - 1]) % 998244353) *
            (func(pos + 1, nflag, nmask) % 998244353)) %
           998244353;
    ans %= 998244353;
    ans += dp2[pos + 1][nflag][nmask] % 998244353;
    ans %= 998244353;
    res += (func(pos + 1, nflag, nmask) % 998244353);
    res %= 998244353;
  }
  dp2[pos][flag][mask] = ans % 998244353;
  return dp1[pos][flag][mask] = res % 998244353;
}
long long init(long long x) {
  memset(dp1, -1, sizeof(dp1));
  memset(dp2, 0, sizeof(dp2));
  num.clear();
  get(x);
  func(0, 0, 0);
  return dp2[0][0][0];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> l >> r >> k;
  pow1[0] = 1;
  for (long long int i = 1; i < 22; i++) {
    pow1[i] = (pow1[i - 1] * 10) % 998244353;
    pow1[i] %= 998244353;
  }
  long long ans1 =
      (init(r) % 998244353 - init(l - 1) % 998244353 + 998244353) % 998244353;
  cout << ans1 % 998244353;
}

### 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;
long long l, r, k;
long long dp1[21][2][1 << 12], dp2[21][2][1 << 12];
std::vector<long long> num;
long long pow1[22];
void get(long long x) {
  while (x) {
    num.push_back(x % 10);
    x /= 10;
  }
  reverse((num).begin(), (num).end());
}
long long func(long long pos, long long flag, long long mask) {
  if (__builtin_popcount(mask) > k) return 0;
  if (pos == num.size()) {
    return 1;
  }
  if (dp1[pos][flag][mask] != -1) return dp1[pos][flag][mask];
  long long res = 0, lm = -1;
  if (!flag) {
    lm = num[pos];
  } else {
    lm = 9;
  }
  long long z;
  if (pos == 0)
    z = 1;
  else
    z = 0;
  long long ans = 0;
  for (long long int i = 0; i < lm + 1; i++) {
    long long nflag = flag;
    if (i < lm && flag == 0) nflag = 1;
    long long nmask = (mask | (1 << i));
    if (mask == 0 && i == 0) nmask = mask;
    ans += (((i * pow1[num.size() - pos - 1]) % 998244353) *
            (func(pos + 1, nflag, nmask) % 998244353)) %
           998244353;
    ans %= 998244353;
    ans += dp2[pos + 1][nflag][nmask] % 998244353;
    ans %= 998244353;
    res += (func(pos + 1, nflag, nmask) % 998244353);
    res %= 998244353;
  }
  dp2[pos][flag][mask] = ans % 998244353;
  return dp1[pos][flag][mask] = res % 998244353;
}
long long init(long long x) {
  memset(dp1, -1, sizeof(dp1));
  memset(dp2, 0, sizeof(dp2));
  num.clear();
  get(x);
  func(0, 0, 0);
  return dp2[0][0][0];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> l >> r >> k;
  pow1[0] = 1;
  for (long long int i = 1; i < 22; i++) {
    pow1[i] = (pow1[i - 1] * 10) % 998244353;
    pow1[i] %= 998244353;
  }
  long long ans1 =
      (init(r) % 998244353 - init(l - 1) % 998244353 + 998244353) % 998244353;
  cout << ans1 % 998244353;
}

```