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 int SIZE = 1 << 10;
int pointer = SIZE;
char buffer[SIZE];
char Advance() {
  if (pointer == SIZE) {
    fread(buffer, 1, SIZE, stdin);
    pointer = 0;
  }
  return buffer[pointer++];
}
int Read() {
  int answer = 0;
  char ch = Advance();
  while (!isdigit(ch)) ch = Advance();
  while (isdigit(ch)) {
    answer = answer * 10 + ch - '0';
    ch = Advance();
  }
  return answer;
}
const int MOD = 998244353;
const int MAXD = 19;
const int SIGMA = 10;
int active[1 + MAXD + 1][1 << SIGMA][2][2], sum[1 + MAXD + 1][1 << SIGMA][2][2];
int power10[1 + MAXD];
int digit[1 + MAXD];
void Add(int &x, int y) {
  x += y;
  if (x >= MOD) x -= MOD;
}
void Subtract(int &x, int y) {
  x -= y;
  if (x < 0) x += MOD;
}
int Count(long long n, int k) {
  for (int i = 1; i <= MAXD; i++) {
    digit[i] = n % 10;
    n /= 10;
  }
  memset(active, 0, sizeof(active));
  memset(sum, 0, sizeof(sum));
  active[MAXD + 1][0][0][0] = 1;
  for (int i = MAXD; i >= 1; i--)
    for (int mask = 0; mask < (1 << SIGMA); mask++)
      for (int nonZero = 0; nonZero < 2; nonZero++)
        for (int smaller = 0; smaller < 2; smaller++)
          for (int d = 0; d < SIGMA; d++)
            if (smaller || d <= digit[i]) {
              int newMask = mask;
              if (nonZero || d != 0) newMask |= (1 << d);
              int newNonZero = nonZero | (d != 0);
              int newSmaller = smaller | (d < digit[i]);
              int value = (1LL * power10[i - 1] * d) % MOD;
              Add(active[i][newMask][newNonZero][newSmaller],
                  active[i + 1][mask][nonZero][smaller]);
              Add(sum[i][newMask][newNonZero][newSmaller],
                  (1LL * value * active[i + 1][mask][nonZero][smaller] +
                   sum[i + 1][mask][nonZero][smaller]) %
                      MOD);
            }
  int answer = 0;
  for (int mask = 0; mask < (1 << SIGMA); mask++)
    if (__builtin_popcount(mask) <= k) Add(answer, sum[1][mask][1][1]);
  return answer;
}
int main() {
  power10[0] = 1;
  for (int i = 1; i <= MAXD; i++)
    power10[i] = (1LL * power10[i - 1] * 10) % MOD;
  long long a, b;
  int k;
  scanf("%I64d%I64d%d", &a, &b, &k);
  int answer = Count(b + 1, k);
  Subtract(answer, Count(a, k));
  printf("%d\n", answer);
  return 0;
}
