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;
inline int Gcd(int X, int Y) { return Y ? Gcd(Y, X % Y) : X; }
inline long long Gcd(long long X, long long Y) { return Y ? Gcd(Y, X % Y) : X; }
const int INF = 0x3f3f3f3f;
int k;
int fu(long long x) {
  if (x <= 9) return x * (x + 1) / 2;
  int num[20], len = 0;
  while (x) num[++len] = x % 10, x /= 10;
  int S = 0;
  long long Sum = 0;
  int f[20][1 << 10], g[20][1 << 10];
  memset(f, 0, sizeof f);
  memset(g, 0, sizeof g);
  for (int i = 1; i <= (num[len] - 1); ++i) {
    f[len][1 << i] = 1;
    g[len][1 << i] = i;
  }
  S = (1 << num[len]);
  Sum = num[len];
  for (int i = len - 1; i >= (1); --i) {
    for (int A = 0; A < (1 << 10); ++A) {
      for (int j = 0; j <= (9); ++j) {
        f[i][A | (1 << j)] += f[i + 1][A];
        f[i][A | (1 << j)] %= 998244353;
        g[i][A | (1 << j)] += (g[i + 1][A] * 10ll % 998244353 +
                               (long long)f[i + 1][A] * j % 998244353) %
                              998244353;
        g[i][A | (1 << j)] %= 998244353;
      }
    }
    for (int j = 0; j <= (num[i] - 1); ++j) {
      ++f[i][S | (1 << j)];
      f[i][S | (1 << j)] %= 998244353;
      g[i][S | (1 << j)] += (Sum * 10ll % 998244353 + j) % 998244353;
      g[i][S | (1 << j)] %= 998244353;
    }
    for (int j = 1; j <= (9); ++j) {
      ++f[i][1 << j];
      f[i][1 << j] %= 998244353;
      g[i][1 << j] += j;
      g[i][1 << j] %= 998244353;
    }
    S |= (1 << num[i]);
    Sum = (Sum * 10ll + num[i]) % 998244353;
  }
  long long Ans = 0;
  for (int A = 0; A < (1 << 10); ++A)
    if (__builtin_popcount(A) <= k) {
      Ans += g[1][A];
      Ans %= 998244353;
    }
  return (Ans + (__builtin_popcount(S) <= k ? Sum : 0)) % 998244353;
}
int main() {
  long long LB, RB;
  scanf("%lld%lld%d", &LB, &RB, &k);
  printf("%d\n", (fu(RB) - fu(LB - 1) + 998244353) % 998244353);
  return 0;
}
