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 long long read() {
  long long x = 0, neg = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') neg = -1;
    c = getchar();
  }
  while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
  return x * neg;
}
long long chong;
inline void print(long long x) {
  if (x < 0) {
    putchar('-');
    print(abs(x));
    return;
  }
  if (x <= 9)
    putchar(x + '0');
  else {
    print(x / 10);
    putchar(x % 10 + '0');
  }
}
inline long long qpow(long long x, long long e, long long _MOD) {
  long long ans = 1;
  while (e) {
    if (e & 1) ans = ans * x % _MOD;
    x = x * x % _MOD;
    e >>= 1;
  }
  return ans;
}
const long long MOD = 998244353;
long long l = read(), r = read(), k = read();
long long a[20], cnt, P[20];
pair<long long, long long> dp[20][1024];
inline pair<long long, long long> dfs(long long pos, long long sta, bool lead,
                                      bool limit) {
  if (!~pos) return pair<long long, long long>(__builtin_popcount(sta) <= k, 0);
  if (!lead && !limit && ~dp[pos][sta].first) return dp[pos][sta];
  long long up = (limit) ? a[pos] : 9;
  pair<long long, long long> ans = pair<long long, long long>(0, 0);
  for (long long i = 0; i <= up; i++) {
    long long nsta = (lead && i == 0) ? sta : (sta | (1 << i));
    pair<long long, long long> nxt =
        dfs(pos - 1, nsta, (lead && i == 0), (limit && i == up));
    ans = pair<long long, long long>(
        (ans.first + nxt.first) % MOD,
        (nxt.first * P[pos] % MOD * i % MOD + nxt.second + ans.second) % MOD);
  }
  if (!limit && !lead) dp[pos][sta] = ans;
  return ans;
}
inline long long solve(long long x) {
  memset(a, 0, sizeof(a));
  cnt = 0;
  while (x) {
    a[cnt++] = x % 10;
    x /= 10;
  }
  for (long long i = 0; i <= 18; i++)
    for (long long j = 0; j <= 1023; j++) dp[i][j].first = dp[i][j].second = -1;
  return dfs(cnt - 1, 0, 1, 1).second;
}
signed main() {
  P[0] = 1;
  for (long long i = 1; i <= 18; i++) P[i] = P[i - 1] * 10 % MOD;
  cout << (solve(r) - solve(l - 1) + MOD) % MOD << endl;
  return 0;
}
