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 int md = 998244353, inf = 1e9 + 7;
const long long hs = 199;
const long double eps = 1e-9, pi = acos(-1);
using namespace std;
int bitcount(int x) {
  int rtn = 0;
  for (int i = 1; i <= x; i <<= 1)
    if (x & i) rtn++;
  return rtn;
}
int S[20][1024][2];
int C[20][1024][2];
long long pw[20];
pair<int, int> solve(const vector<int> &lim, const int &c, int ind, int used,
                     int can) {
  if (bitcount(used) > c) return {0, 0};
  if (ind == lim.size()) return {0, 1};
  if (S[ind][used][can] != -1) return {S[ind][used][can], C[ind][used][can]};
  int &ss = S[ind][used][can];
  int &cc = C[ind][used][can];
  ss = cc = 0;
  pair<int, int> tmp;
  if (can) {
    for (int d = (used ? 0 : 1); d <= 9; d++) {
      tmp = solve(lim, c, ind + 1, used | (1 << d), 1);
      cc = (cc + tmp.second) % md;
      ss = (ss +
            (long long)d * pw[lim.size() - ind - 1] * (long long)tmp.second) %
           md;
      ss = (ss + tmp.first) % md;
    }
  } else {
    for (int d = (used ? 0 : 1); d <= lim[ind]; d++) {
      tmp = solve(lim, c, ind + 1, used | (1 << d), (d < lim[ind] ? 1 : 0));
      cc = (cc + tmp.second) % md;
      ss = (ss +
            (long long)d * pw[lim.size() - ind - 1] * (long long)tmp.second) %
           md;
      ss = (ss + tmp.first) % md;
    }
  }
  return {ss, cc};
}
long long l, r;
int c;
int ans(long long lim) {
  if (lim == 0) return 0;
  for (int i = 0; i < 20; i++)
    for (int j = 0; j < 1024; j++)
      for (int k = 0; k < 2; k++) S[i][j][k] = C[i][j][k] = -1;
  vector<int> s;
  while (lim > 0) {
    s.push_back(lim % 10);
    lim /= 10;
  }
  reverse(s.begin(), s.end());
  int rtn = solve(s, c, 0, 0, 0).first;
  for (int i = 1; i <= s.size(); i++)
    rtn = (rtn + solve(s, c, i, 0, 1).first) % md;
  return rtn;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  pw[0] = 1;
  for (int i = 1; i < 20; i++) pw[i] = 10LL * pw[i - 1] % md;
  cin >> l >> r >> c;
  int a = ans(r), b = ans(l - 1);
  cout << (a - b + md) % md << '\n';
}
