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 = 20, MOD = 998244353;
int k, ans;
long long l, r, pw[N];
string s;
pair<int, int> dp[N][1 << 11][2][2];
void init() {
  pw[0] = 1;
  for (int i = 1; i < N; ++i) {
    pw[i] = (pw[i - 1] * 10LL) % MOD;
  }
}
pair<int, int> solve(int idx = 0, int mask = 0, bool limit = 1, bool zero = 1) {
  if (__builtin_popcount(mask) > k) return {0, 0};
  if (idx >= s.size()) return {1, 0};
  auto& ret = dp[idx][mask][limit][zero];
  if (~ret.first) return ret;
  ret = {0, 0};
  int till = (limit ? s[idx] - '0' : 9);
  for (int i = zero; i <= till; ++i) {
    auto p = solve(idx + 1, mask | (1 << i), limit && (i == till), 0);
    long long val = (i * pw[s.size() - idx - 1]) % MOD;
    ret.first = (ret.first + p.first) % MOD;
    ret.second = (ret.second + p.second) % MOD;
    ret.second = (ret.second + p.first * val) % MOD;
  }
  return ret;
}
int calc(long long x) {
  int ret = 0;
  memset(dp, -1, sizeof(dp));
  s = to_string(x);
  for (int i = 0; i < s.size(); ++i) {
    ret = (ret + solve(i, 0, (i == 0), 1).second) % MOD;
  }
  return ret;
}
int main() {
  cin >> l >> r >> k;
  init();
  ans = ((calc(r) - calc(l - 1)) % MOD + MOD) % MOD;
  cout << ans;
  return 0;
}
