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;
long long int power[20];
pair<long long int, long long int> dp[20][2][2][1024];
vector<long long int> ans;
long long int finalans = 0;
long long int l, r, k;
void init() {
  power[0] = 1;
  for (long long int i = 1; i < 20; i++) {
    power[i] = (10 * power[i - 1]) % 998244353;
  }
  for (long long int i = 0; i < 20; i++) {
    for (long long int j = 0; j < 1024; j++) {
      for (long long int k = 0; k < 2; k++) {
        dp[i][0][k][j] = {-1, 0};
        dp[i][1][k][j] = {-1, 0};
      }
    }
  }
}
vector<long long int> dig(long long int n) {
  vector<long long int> res;
  while (n > 0) {
    res.push_back(n % 10);
    n = n / 10;
  }
  reverse((res).begin(), (res).end());
  return res;
}
pair<long long int, long long int> func(long long int pos, long long int flag,
                                        long long int flag2,
                                        long long int mask) {
  if (pos == ans.size()) {
    if (__builtin_popcountll(mask) <= k) {
      return {1, 0};
    } else {
      return {0, 0};
    }
  }
  if (dp[pos][flag][flag2][mask] != make_pair(-1LL, 0LL)) {
    return dp[pos][flag][flag2][mask];
  }
  long long int up = 9;
  if (flag == 0) up = ans[pos];
  long long int q = ans.size();
  long long int tmp = 0;
  pair<long long int, long long int> anss = {0, 0};
  long long int vall = 0;
  for (long long int i = 0; i <= up; i++) {
    long long int f2 = flag;
    long long int f3 = flag2;
    long long int m2 = mask;
    if (i < up) {
      f2 = 1;
    }
    if (i != 0) {
      f3 = 1;
    }
    m2 = (m2 | (1 << i)) * f3;
    pair<long long int, long long int> ans1 = {0, 0};
    ans1 = func(pos + 1, f2, f3, m2);
    anss.first = (anss.first + ans1.first) % 998244353;
    anss.second =
        (anss.second +
         (ans1.second +
          ((ans1.first) * (power[q - pos - 1] * i) % 998244353) % 998244353) %
             998244353) %
        998244353;
  }
  return dp[pos][flag][flag2][mask] = anss;
}
void solve() {
  init();
  cin >> l >> r >> k;
  ans = dig(r);
  pair<long long int, long long int> t = func(0, 0, 0, 0);
  long long int fval = t.second;
  init();
  ans.clear();
  l--;
  ans = dig(l);
  t = func(0, 0, 0, 0);
  fval = (fval - t.second + 998244353) % 998244353;
  cout << fval << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int tests = 1;
  while (tests--) {
    solve();
  }
}
