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 INF = 0x3f3f3f3f;
pair<long long, long long> pd[1030][20][3];
long long l, r, k, mod = 998244353;
string second;
long long elv[20];
pair<long long, long long> func(int mask, int i, int ig) {
  if (__builtin_popcount(mask) > k) return make_pair(0ll, 0ll);
  if (pd[mask][i][ig].first != -1) return pd[mask][i][ig];
  if (i == second.size()) return make_pair(1ll, 0ll);
  int lim = 9;
  if (ig) lim = second[i] - '0';
  long long ans = 0, jei = 0;
  for (int j = 0; j <= lim; j++) {
    int nova = mask | (1 << j);
    if (nova == 1) nova = 0;
    pair<long long, long long> a = func(nova, i + 1, ig and j == lim);
    ans = (ans + j * (elv[second.size() - i - 1] * a.first % mod) + a.second) %
          mod;
    jei = (jei + a.first) % mod;
  }
  return pd[mask][i][ig] = make_pair(jei, ans);
}
void memseta() {
  for (int i = 0; i < 1030; i++)
    for (int j = 0; j < 20; j++)
      for (int lr = 0; lr < 3; lr++) pd[i][j][lr] = make_pair(-1ll, -1ll);
}
int main() {
  elv[0] = 1;
  for (int i = 1; i < 20; i++) elv[i] = (elv[i - 1] * 10) % mod;
  cin >> l >> r >> k;
  second = to_string(r);
  memseta();
  long long ans = func(0, 0, 1).second;
  second = to_string(l - 1);
  memseta();
  ans = (ans - func(0, 0, 1).second + 3 * mod) % mod;
  cout << ans << endl;
  return 0;
}
