#include <bits/stdc++.h>
using namespace std;
const int N = 20;
const int M = 11;
const int mod = 998244353;
long long int l, r, k, dig, a[N], pw[N];
pair<long long int, long long int> dp[N][1 << M][2][2];
inline long long int md(long long int x) { return x % mod; }
inline long long int add(long long int x, long long int y) {
  return md(md(x) + md(y));
}
inline long long int mul(long long int x, long long int y) {
  return md(md(x) * md(y));
}
pair<long long int, long long int> dfs(int cur, int mask, bool ok, bool st) {
  if (__builtin_popcount(mask) > k) return {0, 0};
  if (cur == dig + 1) return {0, st};
  auto &res = dp[cur][mask][ok][st];
  if (res.first != -1) return res;
  res = {0, 0};
  int big = (ok ? 9 : a[cur]);
  for (int i = big; i >= 0; i--) {
    int addmask = 0;
    if (st || i) addmask = (1 << i);
    auto temp = dfs(cur + 1, mask | addmask, ok || (a[cur] > i), st || i);
    res.first = add(res.first, mul(temp.second, mul(i, pw[dig - cur])));
    res.first = add(res.first, temp.first);
    res.second = add(res.second, temp.second);
  }
  return res;
}
long long int f(long long int x) {
  if (x == 0) return 0;
  for (dig = 0; x;) {
    a[++dig] = x % 10;
    x /= 10;
  }
  reverse(a + 1, a + dig + 1);
  memset(dp, -1, sizeof dp);
  return dfs(1, 0, 0, 0).first;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> l >> r >> k;
  pw[0] = 1;
  for (int i = 1; i <= 20; i++) pw[i] = mul(pw[i - 1], 10);
  cout << add(f(r), mod - f(l - 1));
}

### 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;
const int M = 11;
const int mod = 998244353;
long long int l, r, k, dig, a[N], pw[N];
pair<long long int, long long int> dp[N][1 << M][2][2];
inline long long int md(long long int x) { return x % mod; }
inline long long int add(long long int x, long long int y) {
  return md(md(x) + md(y));
}
inline long long int mul(long long int x, long long int y) {
  return md(md(x) * md(y));
}
pair<long long int, long long int> dfs(int cur, int mask, bool ok, bool st) {
  if (__builtin_popcount(mask) > k) return {0, 0};
  if (cur == dig + 1) return {0, st};
  auto &res = dp[cur][mask][ok][st];
  if (res.first != -1) return res;
  res = {0, 0};
  int big = (ok ? 9 : a[cur]);
  for (int i = big; i >= 0; i--) {
    int addmask = 0;
    if (st || i) addmask = (1 << i);
    auto temp = dfs(cur + 1, mask | addmask, ok || (a[cur] > i), st || i);
    res.first = add(res.first, mul(temp.second, mul(i, pw[dig - cur])));
    res.first = add(res.first, temp.first);
    res.second = add(res.second, temp.second);
  }
  return res;
}
long long int f(long long int x) {
  if (x == 0) return 0;
  for (dig = 0; x;) {
    a[++dig] = x % 10;
    x /= 10;
  }
  reverse(a + 1, a + dig + 1);
  memset(dp, -1, sizeof dp);
  return dfs(1, 0, 0, 0).first;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> l >> r >> k;
  pw[0] = 1;
  for (int i = 1; i <= 20; i++) pw[i] = mul(pw[i - 1], 10);
  cout << add(f(r), mod - f(l - 1));
}

```