#include <bits/stdc++.h>
const long long mod = 998244353;
using namespace std;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
pair<long long, long long> dp[20][2][2][1 << 10][2];
int k;
bool vis[20][2][2][1 << 10][2];
int s1[20], s2[20];
long long pw[20];
pair<long long, long long> solve(int i, bool D, bool U, int mask, bool take) {
  if (i < 0) return {0, 1};
  pair<long long, long long>& ret = dp[i][D][U][mask][take];
  if (vis[i][D][U][mask][take]) return ret;
  vis[i][D][U][mask][take] = 1;
  ret = {0, 0};
  for (int e = 0; e < 10; e++) {
    if ((!D && e > s1[i]) || (!U && e < s2[i])) continue;
    int Mask = mask | ((take || (e > 0)) * (1 << e));
    if (__builtin_popcount(Mask) > k) continue;
    pair<long long, long long> temp =
        solve(i - 1, D || e < s1[i], U || e > s2[i], Mask, take || e > 0);
    ret.first += temp.first;
    ret.first %= mod;
    ret.second += temp.second;
    ret.second %= mod;
    ret.first += temp.second * e % mod * pw[i] % mod;
    ret.first %= mod;
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  pw[0] = 1;
  for (int i = 1; i <= 18; i++) pw[i] = pw[i - 1] * 10LL % mod;
  long long l, r;
  cin >> l >> r >> k;
  for (int i = 0; i < 18; i++) s1[i] = r % 10, r /= 10, s2[i] = l % 10, l /= 10;
  cout << solve(17, 0, 0, 0, 0).first;
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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>
const long long mod = 998244353;
using namespace std;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
pair<long long, long long> dp[20][2][2][1 << 10][2];
int k;
bool vis[20][2][2][1 << 10][2];
int s1[20], s2[20];
long long pw[20];
pair<long long, long long> solve(int i, bool D, bool U, int mask, bool take) {
  if (i < 0) return {0, 1};
  pair<long long, long long>& ret = dp[i][D][U][mask][take];
  if (vis[i][D][U][mask][take]) return ret;
  vis[i][D][U][mask][take] = 1;
  ret = {0, 0};
  for (int e = 0; e < 10; e++) {
    if ((!D && e > s1[i]) || (!U && e < s2[i])) continue;
    int Mask = mask | ((take || (e > 0)) * (1 << e));
    if (__builtin_popcount(Mask) > k) continue;
    pair<long long, long long> temp =
        solve(i - 1, D || e < s1[i], U || e > s2[i], Mask, take || e > 0);
    ret.first += temp.first;
    ret.first %= mod;
    ret.second += temp.second;
    ret.second %= mod;
    ret.first += temp.second * e % mod * pw[i] % mod;
    ret.first %= mod;
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  pw[0] = 1;
  for (int i = 1; i <= 18; i++) pw[i] = pw[i - 1] * 10LL % mod;
  long long l, r;
  cin >> l >> r >> k;
  for (int i = 0; i < 18; i++) s1[i] = r % 10, r /= 10, s2[i] = l % 10, l /= 10;
  cout << solve(17, 0, 0, 0, 0).first;
  return 0;
}

```