#include <bits/stdc++.h>
using namespace std;
const int N = 20, base = 998244353;
long long dem[N][1 << 10][2], sum[N][1 << 10][2];
int q[20];
int getbit(int x, int i) { return x & (1 << i); }
int dembit(int x) {
  int dem = 0;
  for (int i = 0; i <= 9; i++)
    if (getbit(x, i)) {
      dem++;
    }
  return dem;
}
long long kq(long long x, int k) {
  if (x == 0) return 0;
  int top = 0;
  while (x) {
    ++top;
    q[top] = x % 10;
    x /= 10;
  }
  for (int i = 0; i <= 19; i++)
    for (int j = 0; j <= (1 << 10) - 1; j++)
      dem[i][j][0] = dem[i][j][1] = sum[i][j][0] = sum[i][j][1] = 0;
  dem[top + 1][0][1] = 1;
  for (int i = top; i >= 1; i--) {
    int newM;
    for (int chosen = 0; chosen <= 9; chosen++)
      for (int mask = 0; mask <= (1 << 10) - 1; mask++) {
        if (chosen == 0 && mask == 0)
          newM = mask;
        else
          newM = mask | (1 << chosen);
        dem[i][newM][0] = (dem[i][newM][0] + dem[i + 1][mask][0]) % base;
        sum[i][newM][0] = (sum[i][newM][0] + sum[i + 1][mask][0] * 10 +
                           chosen * dem[i + 1][mask][0]) %
                          base;
      }
    int chosen = q[i];
    for (int mask = 0; mask <= (1 << 10) - 1; mask++) {
      if (chosen == 0 && mask == 0)
        newM = mask;
      else
        newM = mask | (1 << chosen);
      dem[i][newM][1] = (dem[i][newM][1] + dem[i + 1][mask][1]) % base;
      sum[i][newM][1] = (sum[i][newM][1] + sum[i + 1][mask][1] * 10 +
                         chosen * dem[i + 1][mask][1]) %
                        base;
    }
    for (int chosen = 0; chosen <= q[i] - 1; chosen++)
      for (int mask = 0; mask <= (1 << 10) - 1; mask++) {
        if (chosen == 0 && mask == 0)
          newM = mask;
        else
          newM = mask | (1 << chosen);
        dem[i][newM][0] = (dem[i][newM][0] + dem[i + 1][mask][1]) % base;
        sum[i][newM][0] = (sum[i][newM][0] + sum[i + 1][mask][1] * 10 +
                           chosen * dem[i + 1][mask][1]) %
                          base;
      }
  }
  long long res = 0;
  for (int mask = 0; mask <= (1 << 10) - 1; mask++)
    if (dembit(mask) <= k) {
      res = (res + sum[1][mask][0] + sum[1][mask][1]) % base;
    }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long l, r;
  int k;
  cin >> l >> r >> k;
  cout << (kq(r, k) - kq(l - 1, k) + 2 * base) % base << endl;
  return 0;
}

### Prompt

Your task is to create 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 int N = 20, base = 998244353;
long long dem[N][1 << 10][2], sum[N][1 << 10][2];
int q[20];
int getbit(int x, int i) { return x & (1 << i); }
int dembit(int x) {
  int dem = 0;
  for (int i = 0; i <= 9; i++)
    if (getbit(x, i)) {
      dem++;
    }
  return dem;
}
long long kq(long long x, int k) {
  if (x == 0) return 0;
  int top = 0;
  while (x) {
    ++top;
    q[top] = x % 10;
    x /= 10;
  }
  for (int i = 0; i <= 19; i++)
    for (int j = 0; j <= (1 << 10) - 1; j++)
      dem[i][j][0] = dem[i][j][1] = sum[i][j][0] = sum[i][j][1] = 0;
  dem[top + 1][0][1] = 1;
  for (int i = top; i >= 1; i--) {
    int newM;
    for (int chosen = 0; chosen <= 9; chosen++)
      for (int mask = 0; mask <= (1 << 10) - 1; mask++) {
        if (chosen == 0 && mask == 0)
          newM = mask;
        else
          newM = mask | (1 << chosen);
        dem[i][newM][0] = (dem[i][newM][0] + dem[i + 1][mask][0]) % base;
        sum[i][newM][0] = (sum[i][newM][0] + sum[i + 1][mask][0] * 10 +
                           chosen * dem[i + 1][mask][0]) %
                          base;
      }
    int chosen = q[i];
    for (int mask = 0; mask <= (1 << 10) - 1; mask++) {
      if (chosen == 0 && mask == 0)
        newM = mask;
      else
        newM = mask | (1 << chosen);
      dem[i][newM][1] = (dem[i][newM][1] + dem[i + 1][mask][1]) % base;
      sum[i][newM][1] = (sum[i][newM][1] + sum[i + 1][mask][1] * 10 +
                         chosen * dem[i + 1][mask][1]) %
                        base;
    }
    for (int chosen = 0; chosen <= q[i] - 1; chosen++)
      for (int mask = 0; mask <= (1 << 10) - 1; mask++) {
        if (chosen == 0 && mask == 0)
          newM = mask;
        else
          newM = mask | (1 << chosen);
        dem[i][newM][0] = (dem[i][newM][0] + dem[i + 1][mask][1]) % base;
        sum[i][newM][0] = (sum[i][newM][0] + sum[i + 1][mask][1] * 10 +
                           chosen * dem[i + 1][mask][1]) %
                          base;
      }
  }
  long long res = 0;
  for (int mask = 0; mask <= (1 << 10) - 1; mask++)
    if (dembit(mask) <= k) {
      res = (res + sum[1][mask][0] + sum[1][mask][1]) % base;
    }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long l, r;
  int k;
  cin >> l >> r >> k;
  cout << (kq(r, k) - kq(l - 1, k) + 2 * base) % base << endl;
  return 0;
}

```