#include <bits/stdc++.h>
using namespace std;
pair<long long, long long> dp[20][(1 << 10) + 2][2][2];
vector<int> v;
int k;
int len;
long long pw[20];
void gen(long long x) {
  v.clear();
  while (x) {
    v.push_back(x % 10);
    x /= 10;
  }
  reverse(v.begin(), v.end());
  len = v.size();
}
pair<long long, long long> fun(int pos, int mask, int z, int neq) {
  if (pos == len) return make_pair(1, 0);
  pair<long long, long long>& ret = dp[pos][mask][z][neq];
  if (ret.first != -1) return ret;
  ret.first = 0;
  ret.second = 0;
  pair<long long, long long> tmp = make_pair(0, 0);
  int cnt = __builtin_popcount(mask);
  if (!z) {
    tmp = fun(pos + 1, mask, z, 1);
  } else if (neq) {
    if ((mask & 1) || cnt < k) tmp = fun(pos + 1, mask | (1 << 0), z, 1);
  } else {
    if (v[pos] == 0) {
      if ((mask & 1) || cnt < k) tmp = fun(pos + 1, mask | (1 << 0), z, 0);
    } else {
      if ((mask & 1) || cnt < k) tmp = fun(pos + 1, mask | (1 << 0), z, 1);
    }
  }
  ret.first += tmp.first;
  ret.second += tmp.second;
  for (int i = 1; i <= 9; i++) {
    if ((mask & (1 << i)) || cnt < k) {
      if (neq) {
        tmp = fun(pos + 1, mask | (1 << i), 1, 1);
      } else if (v[pos] > i) {
        tmp = fun(pos + 1, mask | (1 << i), 1, 1);
      } else if (v[pos] == i) {
        tmp = fun(pos + 1, mask | (1 << i), 1, 0);
      } else {
        continue;
      }
      ret.first += tmp.first;
      ret.first %= 998244353;
      ret.second +=
          (((((long long)i) * pw[len - pos - 1]) % 998244353) * tmp.first) %
              998244353 +
          tmp.second;
      ret.second %= 998244353;
    }
  }
  return ret;
}
int main() {
  pw[0] = 1;
  for (int i = 1; i < 20; i++) {
    pw[i] = pw[i - 1] * 10;
    pw[i] %= 998244353;
  }
  long long a, b;
  cin >> a >> b >> k;
  a--;
  long long A, B;
  memset(dp, -1, sizeof(dp));
  gen(a);
  A = fun(0, 0, 0, 0).second;
  memset(dp, -1, sizeof(dp));
  gen(b);
  B = fun(0, 0, 0, 0).second;
  cout << ((B - A) % 998244353 + 998244353) % 998244353 << endl;
  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;
pair<long long, long long> dp[20][(1 << 10) + 2][2][2];
vector<int> v;
int k;
int len;
long long pw[20];
void gen(long long x) {
  v.clear();
  while (x) {
    v.push_back(x % 10);
    x /= 10;
  }
  reverse(v.begin(), v.end());
  len = v.size();
}
pair<long long, long long> fun(int pos, int mask, int z, int neq) {
  if (pos == len) return make_pair(1, 0);
  pair<long long, long long>& ret = dp[pos][mask][z][neq];
  if (ret.first != -1) return ret;
  ret.first = 0;
  ret.second = 0;
  pair<long long, long long> tmp = make_pair(0, 0);
  int cnt = __builtin_popcount(mask);
  if (!z) {
    tmp = fun(pos + 1, mask, z, 1);
  } else if (neq) {
    if ((mask & 1) || cnt < k) tmp = fun(pos + 1, mask | (1 << 0), z, 1);
  } else {
    if (v[pos] == 0) {
      if ((mask & 1) || cnt < k) tmp = fun(pos + 1, mask | (1 << 0), z, 0);
    } else {
      if ((mask & 1) || cnt < k) tmp = fun(pos + 1, mask | (1 << 0), z, 1);
    }
  }
  ret.first += tmp.first;
  ret.second += tmp.second;
  for (int i = 1; i <= 9; i++) {
    if ((mask & (1 << i)) || cnt < k) {
      if (neq) {
        tmp = fun(pos + 1, mask | (1 << i), 1, 1);
      } else if (v[pos] > i) {
        tmp = fun(pos + 1, mask | (1 << i), 1, 1);
      } else if (v[pos] == i) {
        tmp = fun(pos + 1, mask | (1 << i), 1, 0);
      } else {
        continue;
      }
      ret.first += tmp.first;
      ret.first %= 998244353;
      ret.second +=
          (((((long long)i) * pw[len - pos - 1]) % 998244353) * tmp.first) %
              998244353 +
          tmp.second;
      ret.second %= 998244353;
    }
  }
  return ret;
}
int main() {
  pw[0] = 1;
  for (int i = 1; i < 20; i++) {
    pw[i] = pw[i - 1] * 10;
    pw[i] %= 998244353;
  }
  long long a, b;
  cin >> a >> b >> k;
  a--;
  long long A, B;
  memset(dp, -1, sizeof(dp));
  gen(a);
  A = fun(0, 0, 0, 0).second;
  memset(dp, -1, sizeof(dp));
  gen(b);
  B = fun(0, 0, 0, 0).second;
  cout << ((B - A) % 998244353 + 998244353) % 998244353 << endl;
  return 0;
}

```