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.
#include <bits/stdc++.h>
using namespace std;
const long long int modulo = 998244353;
void add(long long int &a, long long int b) { a = (a + b) % modulo; }
long long int expo[11][25];
int n;
long long int digit[25];
long long int numbefore(long long int r, int c) {
  long long int(&ex)[25] = expo[__builtin_popcount(c)];
  n = 0;
  for (; r; r /= 10) digit[n++] = r % 10;
  reverse(digit, digit + n);
  long long int sol = 0;
  long long int acum = 0;
  if ((c & 1) == 0)
    for (int i = 1; i < n; i++)
      for (int pos = i; pos < n; pos++)
        for (int d = 0; d < 10; d++)
          if ((c >> d) & 1)
            add(sol, d * expo[10][n - 1 - pos] % modulo * ex[n - 1 - i]);
  for (int i = 0; i < n; i++) {
    long long int under = 0;
    for (long long int d = 0; d < digit[i]; d++) {
      if ((c >> d) & 1) {
        under++;
        add(sol, d * expo[10][n - 1 - i] % modulo * ex[n - 1 - i]);
      }
    }
    add(sol, acum * under % modulo * ex[n - 1 - i]);
    for (long long int pos = n - 1; pos > i; pos--)
      for (long long int d = 0; d < 10; d++)
        if ((c >> d) & 1)
          add(sol, d * expo[10][n - 1 - pos] % modulo * under % modulo *
                       ex[n - 1 - i - 1]);
    if (((c >> digit[i]) & 1) == 0) return sol;
    add(acum, digit[i] * expo[10][n - 1 - i]);
  }
  return sol;
}
int k;
long long int memo[1 << 10];
long long int compute(long long int r) {
  long long int sol = 0;
  for (int c = 1; c < 1 << 10; c++) {
    if (__builtin_popcount(c) <= k) {
      memo[c] = numbefore(r, c);
      for (int d = 1; d < c; d++)
        if ((d & c) == d) add(memo[c], -memo[d]);
      add(sol, memo[c]);
    }
  }
  return sol;
}
int main() {
  for (int b = 0; b < 11; b++) {
    expo[b][0] = 1;
    for (int e = 1; e < 25; e++) expo[b][e] = expo[b][e - 1] * b % modulo;
  }
  long long int l, r;
  cin >> l >> r >> k;
  long long int sol = compute(r + 1) - compute(l);
  sol = (sol % modulo + modulo) % modulo;
  cout << sol << endl;
}
