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 long long mod = 998244353;
map<tuple<unsigned long, int, int>, pair<long long, long long> > m;
pair<long long, long long> zml(long long x, bitset<10> s, int p, int k,
                               bool limit) {
  if (int(s.count()) > k) return {0, 0};
  if (p == -1) return {1, 0};
  if (s.count() == 1 && s[0]) s.reset();
  if (!limit && m.count({s.to_ulong(), p, k})) return m[{s.to_ulong(), p, k}];
  long long r = 1;
  for (int i = 0; i < p; i++) r *= 10;
  long long y = x / r;
  long long ret1 = 0;
  long long ret2 = 0;
  for (int i = 0; i <= y; i++) {
    bitset<10> t = s;
    t[i] = 1;
    pair<long long, long long> z =
        zml(i == y ? x - y * r : r - 1, t, p - 1, k, limit && i == y);
    ret1 += z.first;
    ret1 %= mod;
    ret2 += z.second;
    ret2 %= mod;
    ret2 += z.first * (((r % mod) * i) % mod);
    ret2 %= mod;
  }
  if (!limit)
    return m[{s.to_ulong(), p, k}] = {ret1, ret2};
  else
    return {ret1, ret2};
}
int main() {
  long long l, r;
  int k;
  cin >> l >> r >> k;
  cout << (zml(r, bitset<10>(), 18, k, 1).second -
           zml(l - 1, bitset<10>(), 18, k, 1).second + mod) %
              mod
       << endl;
}
