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;
string l, r;
long long k;
long long cache1[20][1026][2];
long long cache[20][1026][2];
long long power[20];
long long dp1(long long pos, long long bit, long long f) {
  if (__builtin_popcountll(bit) > k) return cache1[pos][bit][f] = 0;
  if (pos == r.length()) return cache1[pos][bit][f] = 1;
  if (cache1[pos][bit][f] != -1) return cache1[pos][bit][f];
  long long ans = 0;
  for (long long i = 0; i <= (f ? 9 : r[pos] - '0'); i++)
    if (bit == 0 && i == 0)
      ans += dp1(pos + 1, bit, 1);
    else
      ans += dp1(pos + 1, bit | (1 << i), f | ((r[pos] - '0') != i));
  ans %= mod;
  return cache1[pos][bit][f] = ans;
}
long long dp2(long long pos, long long bit, long long f) {
  if (__builtin_popcountll(bit) > k) return 0;
  if (pos == r.length()) return 0;
  if (cache[pos][bit][f] != -1) return cache[pos][bit][f];
  long long ans = 0;
  for (long long i = 0; i <= (f ? 9 : r[pos] - '0'); i++) {
    long long p = (i * power[r.length() - pos - 1]) % mod;
    if (bit == 0 && i == 0)
      ans += dp2(pos + 1, bit, 1) + cache1[pos + 1][bit][1] * p % mod;
    else
      ans +=
          dp2(pos + 1, bit | (1 << i), f | ((r[pos] - '0') != i)) +
          cache1[pos + 1][bit | (1 << i)][f | ((r[pos] - '0') != i)] * p % mod;
    ans %= mod;
  }
  return cache[pos][bit][f] = ans;
}
long long solve(string a) {
  r = a;
  memset(cache1, -1, sizeof(cache1));
  dp1(0, 0, 0);
  memset(cache, -1, sizeof(cache));
  return dp2(0, 0, 0);
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> l >> r;
  cin >> k;
  power[0] = 1;
  for (long long i = 1; i < 20; i++) power[i] = (power[i - 1] * 10) % mod;
  long long ans = (solve(r) - solve(l) + mod) % mod;
  long long ch = 0;
  long long val = 0;
  for (long long i = 0; i < l.length(); i++)
    ch |= (1 << (l[i] - 48)), val = val * 10 + l[i] - 48;
  if (__builtin_popcountll(ch) <= k) ans += val;
  cout << ans % mod;
}
