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 N = 1e6 + 5, mod = 998244353;
int a[20];
int k;
long long p[25];
pair<long long, long long> dp[20][1 << 15];
pair<long long, long long> dfs(int pos, int statu, bool zero, bool limit) {
  if (!pos) return make_pair(__builtin_popcount(statu) <= k, 0);
  if (!limit && !zero && dp[pos][statu].first != -1) return dp[pos][statu];
  int up = limit ? a[pos] : 9;
  pair<long long, long long> ans = make_pair(0, 0);
  for (int i = 0; i <= up; i++) {
    pair<long long, long long> res;
    if (zero && i == 0)
      res = dfs(pos - 1, statu, zero, limit && i == up);
    else
      res = dfs(pos - 1, statu | (1 << i), zero && i == 0, limit && i == up);
    ans.first = (ans.first + res.first) % mod;
    ans.second =
        (ans.second + res.second + i * res.first % mod * p[pos - 1] % mod) %
        mod;
  }
  if (!limit && !zero) dp[pos][statu] = ans;
  return ans;
}
long long cal(long long x) {
  int tot = 0;
  while (x) {
    a[++tot] = x % 10;
    x = x / 10;
  }
  memset(dp, -1, sizeof(dp));
  return dfs(tot, 0, 1, 1).second;
}
void solve() {
  long long l, r;
  p[0] = 1LL;
  for (int i = 1; i <= 20; i++) p[i] = (p[i - 1] * 10LL) % mod;
  scanf("%lld %lld %d", &l, &r, &k);
  long long ans = (cal(r) - cal(l - 1) + mod) % mod;
  printf("%lld\n", ans);
}
int main() {
  solve();
  return 0;
}
