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;
long long cache1[20][1024][2];
long long cache2[20][1024][2];
long long L, R, k, n;
vector<long long> digits;
long long pow(long long a, long long b, long long p) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % p;
    b /= 2;
    a = (a * a) % p;
  }
  return res;
}
void getDigits(long long x) {
  digits.clear();
  while (x) {
    digits.push_back(x % 10);
    x /= 10;
  }
  reverse(digits.begin(), digits.end());
  n = digits.size();
}
long long countSetBits(long long &x) { return __builtin_popcount(x); }
long long dp1(long long idx, long long mask, long long tight) {
  if (countSetBits(mask) > k) return 0;
  if (idx == n) return 1;
  long long &res = cache1[idx][mask][tight];
  if (res != -1) return res;
  res = 0;
  long long limit = tight ? digits[idx] : 9;
  for (long long i = 0; i <= limit; i++) {
    long long newTight = (i == limit) ? tight : 0;
    res += dp1(idx + 1, mask | (1 << i), newTight);
    res %= MOD;
  }
  return res;
}
long long dp2(long long idx, long long mask, long long tight,
              long long hasStarted) {
  if (countSetBits(mask) > k) return 0;
  if (idx == n) return 0;
  long long &res = cache2[idx][mask][tight];
  if (res != -1) return res;
  res = 0;
  long long limit = tight ? digits[idx] : 9;
  if (hasStarted) {
    for (long long i = 0; i <= limit; i++) {
      long long newTight = (i == limit) ? tight : 0;
      long long ways = dp1(idx + 1, mask | (1 << i), newTight);
      long long val = (i * pow(10, n - idx - 1, MOD)) % MOD;
      val *= ways;
      val %= MOD;
      res += val;
      res += dp2(idx + 1, mask | (1 << i), newTight, hasStarted);
      res %= MOD;
    }
  } else {
    for (long long i = 1; i <= limit; i++) {
      long long newTight = (i == limit) ? tight : 0;
      long long ways = dp1(idx + 1, mask | (1 << i), newTight);
      long long val = (i * pow(10, n - idx - 1, MOD)) % MOD;
      val *= ways;
      val %= MOD;
      res += val;
      res += dp2(idx + 1, mask | (1 << i), newTight, 1);
      res %= MOD;
    }
    res += dp2(idx + 1, mask, 0, 0);
    res %= MOD;
  }
  return res;
}
long long f(long long x) {
  memset(cache1, -1, sizeof(cache1));
  memset(cache2, -1, sizeof(cache2));
  getDigits(x);
  long long ans = dp2(0, 0, 1, 0);
  return ans;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> L >> R >> k;
  if (L > R) swap(L, R);
  long long ans = f(R) - f(L - 1) + MOD;
  ans %= MOD;
  cout << ans << "\n";
  return 0;
}
