#include <bits/stdc++.h>
using namespace std;
string l, r, s;
long long int k;
long long int p10[25];
pair<long long int, long long int> dp[2][25];
long long int ann[1 << 13];
bool cool[11];
long long int tr;
pair<long long int, long long int> rec(long long int i, long long int z) {
  if (i >= (int)s.length()) return {0, 1};
  if (dp[z][i] != make_pair(-1ll, -1ll)) return dp[z][i];
  pair<long long int, long long int> an = {0, 0};
  for (long long int h = (i == tr ? 1 : 0); h <= (z ? 9 : s[i] - '0'); h++) {
    if (cool[h]) {
      pair<long long int, long long int> d = rec(i + 1, z | (h < s[i] - '0'));
      an.second += d.second;
      an.first += ((p10[(int)s.length() - 1 - i] * h) % 998244353) * d.second %
                  998244353;
      an.first += d.first;
      an.first %= 998244353;
      an.second %= 998244353;
    }
  }
  return dp[z][i] = an;
}
void work(long long int j) {
  memset(cool, 0, sizeof cool);
  for (int i = 0; i <= 9; i++)
    if ((j & (1 << i))) cool[i] = 1;
  s = l;
  long long int x = 0;
  for (int i = 0; i < s.length(); i++) {
    tr = i;
    for (int i = 0; i <= 20; i++)
      for (int j = 0; j < 2; j++) dp[j][i] = {-1, -1};
    x += rec(i, (i > 0)).first;
    x %= 998244353;
  }
  s = r;
  long long int y = 0;
  for (int i = 0; i < s.length(); i++) {
    tr = i;
    for (int i = 0; i <= 20; i++)
      for (int j = 0; j < 2; j++) dp[j][i] = {-1, -1};
    y += rec(i, (i > 0)).first;
    y %= 998244353;
  }
  ann[j] = (y - x + 998244353) % 998244353;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> l >> r;
  cin >> k;
  p10[0] = 1;
  for (int i = 1; i <= 24; i++) p10[i] = (p10[i - 1] * 10ll) % 998244353;
  for (int i = 1; i < (1 << 10); i++) work(i);
  long long int ans = 0;
  for (int i = 1; i < (1 << 10); i++) {
    for (int j = i; j > 0; j = (j - 1) & i) {
      if (j != i) ann[i] -= ann[j];
    }
  }
  for (int i = 1; i < (1 << 10); i++) {
    long long int j = __builtin_popcount(i);
    if (j <= k) ans += ann[i];
    ans %= 998244353;
    if (ans < 0) ans += 998244353;
  }
  int j = 0;
  for (auto h : l) j |= (1 << (h - '0'));
  if (__builtin_popcount(j) <= k) {
    long long int cur = 0;
    for (auto h : l) cur = cur * 10 + h - '0';
    ans += cur;
    ans %= 998244353;
  }
  if (ans < 0) ans += 998244353;
  cout << ans << '\n';
  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;
string l, r, s;
long long int k;
long long int p10[25];
pair<long long int, long long int> dp[2][25];
long long int ann[1 << 13];
bool cool[11];
long long int tr;
pair<long long int, long long int> rec(long long int i, long long int z) {
  if (i >= (int)s.length()) return {0, 1};
  if (dp[z][i] != make_pair(-1ll, -1ll)) return dp[z][i];
  pair<long long int, long long int> an = {0, 0};
  for (long long int h = (i == tr ? 1 : 0); h <= (z ? 9 : s[i] - '0'); h++) {
    if (cool[h]) {
      pair<long long int, long long int> d = rec(i + 1, z | (h < s[i] - '0'));
      an.second += d.second;
      an.first += ((p10[(int)s.length() - 1 - i] * h) % 998244353) * d.second %
                  998244353;
      an.first += d.first;
      an.first %= 998244353;
      an.second %= 998244353;
    }
  }
  return dp[z][i] = an;
}
void work(long long int j) {
  memset(cool, 0, sizeof cool);
  for (int i = 0; i <= 9; i++)
    if ((j & (1 << i))) cool[i] = 1;
  s = l;
  long long int x = 0;
  for (int i = 0; i < s.length(); i++) {
    tr = i;
    for (int i = 0; i <= 20; i++)
      for (int j = 0; j < 2; j++) dp[j][i] = {-1, -1};
    x += rec(i, (i > 0)).first;
    x %= 998244353;
  }
  s = r;
  long long int y = 0;
  for (int i = 0; i < s.length(); i++) {
    tr = i;
    for (int i = 0; i <= 20; i++)
      for (int j = 0; j < 2; j++) dp[j][i] = {-1, -1};
    y += rec(i, (i > 0)).first;
    y %= 998244353;
  }
  ann[j] = (y - x + 998244353) % 998244353;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> l >> r;
  cin >> k;
  p10[0] = 1;
  for (int i = 1; i <= 24; i++) p10[i] = (p10[i - 1] * 10ll) % 998244353;
  for (int i = 1; i < (1 << 10); i++) work(i);
  long long int ans = 0;
  for (int i = 1; i < (1 << 10); i++) {
    for (int j = i; j > 0; j = (j - 1) & i) {
      if (j != i) ann[i] -= ann[j];
    }
  }
  for (int i = 1; i < (1 << 10); i++) {
    long long int j = __builtin_popcount(i);
    if (j <= k) ans += ann[i];
    ans %= 998244353;
    if (ans < 0) ans += 998244353;
  }
  int j = 0;
  for (auto h : l) j |= (1 << (h - '0'));
  if (__builtin_popcount(j) <= k) {
    long long int cur = 0;
    for (auto h : l) cur = cur * 10 + h - '0';
    ans += cur;
    ans %= 998244353;
  }
  if (ans < 0) ans += 998244353;
  cout << ans << '\n';
  return 0;
}

```