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[20][1 << 10];
int a[20];
long long pw[20];
int change(long long n) {
  int cnt = 0;
  while (n) {
    if (n & 1) cnt++;
    n /= 2;
  }
  return cnt;
}
int k;
long long pow(int e, int n) {
  long long ans = 1;
  while (n--) {
    ans *= e;
  }
  return ans;
}
pair<long long, long long> dfs(int len, long long Set, bool zero, bool limit) {
  if (len == 0) {
    if (change(Set) <= k) return make_pair(1, 0);
    return make_pair(0, 0);
  }
  if (!limit && dp[len][Set].first != 0 && !zero) return dp[len][Set];
  int u = limit ? a[len - 1] : 9;
  pair<long long, long long> ans = make_pair(0, 0);
  for (int i = 0; i <= u; i++) {
    pair<long long, long long> tem = make_pair(0, 0);
    if (change(Set | (1 << i)) > k) continue;
    if (zero && i == 0)
      tem = dfs(len - 1, Set, zero, i == u && limit);
    else {
      tem = dfs(len - 1, Set | (1 << i), zero && i == 0, i == u && limit);
    }
    ans.first += tem.first;
    ans.first %= mod;
    ans.second =
        (ans.second + i * tem.first % mod * pw[len - 1] % mod + tem.second) %
        mod;
  }
  if (!limit && !zero) dp[len][Set] = ans;
  return ans;
}
void init() {
  pw[0] = 1;
  for (int i = 1; i < 20; i++) pw[i] = (pw[i - 1] * 10) % mod;
  for (int i = 0; i < 20; i++)
    for (int j = 0; j < 1 << 9; j++) dp[i][j].first = dp[i][j].second = 0;
}
long long solve(long long n) {
  int len = 0;
  init();
  while (n) {
    a[len++] = n % 10;
    n /= 10;
  }
  return dfs(len, 0, 1, 1).second;
}
int main() {
  long long l, r;
  while (cin >> l >> r >> k) {
    cout << (solve(r) - solve(l - 1) + mod) % mod << endl;
  }
  return 0;
}
