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>
const long long N = 20, MOD = 998244353ll;
struct Node {
  long long tot;
  long long sum;
};
long long f[N][(1 << 12)][12][2][2];
long long tot[N][(1 << 12)][12][2][2];
long long a[N], len;
long long Pow[N];
long long L, R;
long long k;
inline void make(long long x) {
  len = 0;
  while (x) {
    a[++len] = x % 10;
    x /= 10;
  }
  return;
}
Node dfs(long long i, long long state, long long cnt, long long last, bool zero,
         bool high) {
  if (cnt > k) {
    return (Node){tot[i][state][last][zero][high] = 0,
                  f[i][state][last][zero][high] = 0};
  }
  if (!i) {
    return (Node){tot[i][state][last][zero][high] = 1,
                  f[i][state][last][zero][high] = 0};
  }
  if (~f[i][state][last][zero][high]) {
    return (Node){tot[i][state][last][zero][high] % MOD,
                  f[i][state][last][zero][high] % MOD};
  }
  long long totres = 0, sumres = 0;
  for (register long long j = 0; j <= 9; ++j) {
    if (high && j > a[i]) break;
    Node result;
    if (zero) {
      if (!j)
        result = dfs(i - 1, state, cnt, j, zero, high && j == a[i]);
      else
        result =
            dfs(i - 1, state | (1 << j), cnt + 1, j, false, high && j == a[i]);
    } else {
      if ((state | (1 << j)) != state)
        result =
            dfs(i - 1, state | (1 << j), cnt + 1, j, zero, high && j == a[i]);
      else
        result = dfs(i - 1, state, cnt, j, zero, high && j == a[i]);
    }
    totres = (totres + result.tot) % MOD;
    sumres = (sumres + (result.tot * j % MOD * Pow[i] + result.sum)) % MOD;
  }
  return (Node){tot[i][state][last][zero][high] = totres,
                f[i][state][last][zero][high] = sumres};
}
inline long long work(long long x) {
  make(x);
  memset(f, -1, sizeof(f));
  memset(tot, -1, sizeof(tot));
  return dfs(len, 0, 0, 0, true, true).sum;
}
signed main(void) {
  scanf("%I64d%I64d%I64d", &L, &R, &k);
  Pow[1] = 1;
  for (register long long i = 2; i <= 19; ++i) {
    Pow[i] = (Pow[i - 1] * 10ll) % MOD;
  }
  printf("%I64d", (work(R) + MOD - work(L - 1)) % MOD);
  return 0;
}
