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;
template <class T>
using VV = vector<vector<T>>;
template <class T>
inline T SMIN(T& a, const T b) {
  return a = (a > b) ? b : a;
}
template <class T>
inline T SMAX(T& a, const T b) {
  return a = (a < b) ? b : a;
}
template <class Iter>
void dumpc(Iter begin, Iter end) {
  if (true) {
    for (; begin != end; ++begin) {
      cerr << *begin << ',';
    }
    cerr << endl;
  }
}
const long long MOD = 998244353;
long long dp[19][1 << 10][2], dps[19][1 << 10][2];
long long L, R;
int K;
long long P[20];
long long f(long long AA) {
  vector<int> A;
  while (AA > 0) A.push_back(AA % 10), AA /= 10LL;
  reverse((A).begin(), (A).end());
  memset(dp, 0, sizeof(dp)), memset(dps, 0, sizeof(dps));
  int N = int((A).size());
  for (int i = 0; i < (N); ++i) dp[i][0][i > 0] = 1;
  for (int i = 0; i < (N); ++i)
    for (int mask = 0; mask < (1 << 10); ++mask)
      for (int less = 0; less < (2); ++less) {
        int ld = mask > 0 ? 0 : 1;
        int ud = less ? 10 : A[i] + 1;
        for (int d = (ld); d < (ud); ++d) {
          int mask2 = mask | (1 << d), less2 = less | (d < A[i]);
          (dp[i + 1][mask2][less2] += dp[i][mask][less]) %= MOD;
          (dps[i + 1][mask2][less2] +=
           dps[i][mask][less] +
           dp[i][mask][less] * P[N - i - 1] % MOD * d % MOD) %= MOD;
        }
      }
  long long res = 0;
  for (int mask = 0; mask < (1 << 10); ++mask)
    for (int less = 0; less < (2); ++less)
      if (__builtin_popcount(mask) <= K) {
        (res += dps[N][mask][less]) %= MOD;
      }
  return res;
}
void solve() {
  P[0] = 1;
  for (int i = 0; i < (18); ++i) P[i + 1] = (P[i] * 10) % MOD;
  long long r = f(R), l = f(L - 1);
  cout << (MOD + r - l) % MOD << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> L >> R >> K;
  solve();
  return 0;
}
