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 l, r, k;
int n;
vector<long long> vec;
long long p10[21];
pair<long long, long long> dp[20][(1 << 11)][2];
pair<long long, long long> call(int pos, int mask, int eq) {
  if (pos == n) {
    if (__builtin_popcount(mask) > k) return make_pair(0, 0);
    return make_pair(1, 0);
  }
  if (dp[pos][mask][eq].first != -1) return dp[pos][mask][eq];
  pair<long long, long long> ret = make_pair(0, 0), p;
  if (eq) {
    for (long long i = 0; i < vec[pos]; i++) {
      p = call(pos + 1, (mask | (1 << i)), 0);
      ret.first += p.first;
      ret.second +=
          p.second +
          (p.first * ((i * p10[n - pos - 1]) % 998244353LL)) % 998244353LL;
    }
    p = call(pos + 1, (mask | (1 << vec[pos])), 1);
    ret.first += p.first;
    ret.second +=
        p.second +
        (p.first * ((vec[pos] * p10[n - pos - 1]) % 998244353LL)) % 998244353LL;
  } else {
    for (long long i = 0; i < 10; i++) {
      p = call(pos + 1, (mask | (1 << i)), 0);
      ret.first += p.first;
      ret.second +=
          p.second +
          (p.first * ((i * p10[n - pos - 1]) % 998244353LL)) % 998244353LL;
    }
  }
  ret.first %= 998244353LL;
  ret.second %= 998244353LL;
  return dp[pos][mask][eq] = ret;
}
long long calc(long long num) {
  if (num == 0) return 0;
  vec.clear();
  while (num > 0) {
    vec.push_back(num % 10);
    num /= 10;
  }
  reverse(vec.begin(), vec.end());
  n = vec.size();
  memset(dp, -1, sizeof dp);
  long long ret = 0;
  pair<long long, long long> p;
  for (long long i = 1; i < vec[0]; i++) {
    p = call(1, (1 << i), 0);
    ret +=
        p.second + (p.first * ((i * p10[n - 1]) % 998244353LL)) % 998244353LL;
  }
  p = call(1, (1 << vec[0]), 1);
  ret += p.second +
         (p.first * ((vec[0] * p10[n - 1]) % 998244353LL)) % 998244353LL;
  for (long long i = 1; i < n; i++) {
    for (long long j = 1; j <= 9; j++) {
      p = call(i + 1, (1 << j), 0);
      ret += p.second +
             (p.first * ((j * p10[n - i - 1]) % 998244353LL)) % 998244353LL;
    }
  }
  return ret % 998244353LL;
}
int main() {
  p10[0] = 1;
  for (int i = 1; i <= 20; i++) p10[i] = (p10[i - 1] * 10LL) % 998244353LL;
  cin >> l >> r >> k;
  cout << ((calc(r) - calc(l - 1) + 998244353LL) % 998244353LL) << endl;
  return 0;
}
