#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
int k;
int c[25];
struct Point {
  long long cnt, sum;
} dp[20][1025][20], temp;
long long pre[25];
Point dfs(int pos, int state, int xxx, bool lead, bool limit) {
  if (state > k) {
    temp.cnt = temp.sum = 0;
    return temp;
  }
  if (pos == 0) {
    temp.cnt = 1;
    temp.sum = 0;
    return temp;
  }
  if (!limit && !lead && dp[pos][xxx][state].cnt != -1) {
    return dp[pos][xxx][state];
  }
  int up = (limit ? c[pos] : 9);
  long long cnt = 0, sum = 0;
  for (int i = 0; i <= up; i++) {
    if (xxx & (1 << i)) {
      temp = dfs(pos - 1, state, xxx, lead && i == 0, limit && i == c[pos]);
      cnt += temp.cnt;
      sum += (temp.sum + temp.cnt * i % mod * pre[pos] % mod) % mod;
      sum %= mod;
    } else {
      if (lead == 1 && i == 0) {
        temp = dfs(pos - 1, state, xxx, 1, limit && i == c[pos]);
        cnt += temp.cnt;
        sum += temp.sum;
        sum %= mod;
      } else {
        temp = dfs(pos - 1, state + 1, xxx ^ (1 << i), 0, limit && i == c[pos]);
        cnt += temp.cnt;
        sum += (temp.sum + temp.cnt * i % mod * pre[pos] % mod) % mod;
        sum %= mod;
      }
    }
  }
  temp.cnt = cnt;
  temp.sum = sum;
  if (lead == 0 && limit == 0) dp[pos][xxx][state] = temp;
  return temp;
}
Point solve(long long n) {
  int cnt = 0;
  while (n) {
    c[++cnt] = n % 10;
    n /= 10;
  }
  return dfs(cnt, 0, 0, true, true);
}
int main() {
  long long a, b;
  memset(dp, -1, sizeof dp);
  cin >> a >> b >> k;
  pre[1] = 1;
  for (int i = 2; i <= 24; i++) pre[i] = pre[i - 1] * 10 % mod;
  cout << (solve(b).sum - solve(a - 1).sum + mod) % mod << 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 long long mod = 998244353;
int k;
int c[25];
struct Point {
  long long cnt, sum;
} dp[20][1025][20], temp;
long long pre[25];
Point dfs(int pos, int state, int xxx, bool lead, bool limit) {
  if (state > k) {
    temp.cnt = temp.sum = 0;
    return temp;
  }
  if (pos == 0) {
    temp.cnt = 1;
    temp.sum = 0;
    return temp;
  }
  if (!limit && !lead && dp[pos][xxx][state].cnt != -1) {
    return dp[pos][xxx][state];
  }
  int up = (limit ? c[pos] : 9);
  long long cnt = 0, sum = 0;
  for (int i = 0; i <= up; i++) {
    if (xxx & (1 << i)) {
      temp = dfs(pos - 1, state, xxx, lead && i == 0, limit && i == c[pos]);
      cnt += temp.cnt;
      sum += (temp.sum + temp.cnt * i % mod * pre[pos] % mod) % mod;
      sum %= mod;
    } else {
      if (lead == 1 && i == 0) {
        temp = dfs(pos - 1, state, xxx, 1, limit && i == c[pos]);
        cnt += temp.cnt;
        sum += temp.sum;
        sum %= mod;
      } else {
        temp = dfs(pos - 1, state + 1, xxx ^ (1 << i), 0, limit && i == c[pos]);
        cnt += temp.cnt;
        sum += (temp.sum + temp.cnt * i % mod * pre[pos] % mod) % mod;
        sum %= mod;
      }
    }
  }
  temp.cnt = cnt;
  temp.sum = sum;
  if (lead == 0 && limit == 0) dp[pos][xxx][state] = temp;
  return temp;
}
Point solve(long long n) {
  int cnt = 0;
  while (n) {
    c[++cnt] = n % 10;
    n /= 10;
  }
  return dfs(cnt, 0, 0, true, true);
}
int main() {
  long long a, b;
  memset(dp, -1, sizeof dp);
  cin >> a >> b >> k;
  pre[1] = 1;
  for (int i = 2; i <= 24; i++) pre[i] = pre[i - 1] * 10 % mod;
  cout << (solve(b).sum - solve(a - 1).sum + mod) % mod << endl;
  return 0;
}

```