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;
pair<long long, long long> dp[19][1024];
int a[19];
int d;
long long pw[20];
long long n;
void update(long long &a, long long b) { a = (a + b) % mod; }
pair<long long, long long> dfs(int len, int ok, bool flag, bool zero) {
  if (len == -1) {
    return make_pair(1, 0);
  }
  if (!flag && !zero && dp[len][ok].first != -1) return dp[len][ok];
  pair<long long, long long> ans = make_pair(0, 0);
  int limit = flag ? a[len] : 9;
  for (int i = 0; i <= limit; ++i) {
    int mask = ok | (1 << i);
    if (i == 0 && zero) mask = ok;
    if (__builtin_popcount(mask) > d) continue;
    pair<long long, long long> to =
        dfs(len - 1, mask, flag && i == a[len], zero && i == 0);
    update(ans.first, to.first);
    update(ans.second, to.second);
    update(ans.second, 1LL * to.first * pw[len] % mod * i % mod);
  }
  if (!flag) dp[len][ok] = ans;
  return ans;
}
long long cal(long long n) {
  int len = -1;
  while (n) {
    a[++len] = n % 10;
    n /= 10;
  }
  for (int i = 0; i < 19; ++i)
    for (int j = 0; j < 1024; ++j) dp[i][j] = make_pair(-1, -1);
  return dfs(len, 0, 1, 1).second;
}
int main() {
  pw[0] = 1;
  for (int i = 1; i < 20; ++i) pw[i] = pw[i - 1] * 10 % mod;
  long long l, r;
  cin >> l >> r >> d;
  printf("%lld\n", (cal(r) - cal(l - 1) + mod) % mod);
  return 0;
}
