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 int mod = 998244353;
int k;
long long int PW[20];
int SUM[20][2][1 << 10];
int C[20][2][1 << 10];
string str(long long int x) {
  string res;
  while (x) {
    res += x % 10 + '0';
    x /= 10;
  }
  reverse(res.begin(), res.end());
  return res;
}
int cnt(const string &s, int i, int small, int mask) {
  if (i == s.size()) {
    return 1;
  }
  int &res = C[i][small][mask];
  if (res != -1) {
    return res;
  }
  res = 0;
  for (int d = 0; d < 10; d++) {
    if (!small && d > s[i] - '0') {
      continue;
    }
    int nwMask = mask | (1 << d);
    if (!mask && !d) nwMask = 0;
    if (__builtin_popcount(nwMask) > k) {
      continue;
    }
    res += cnt(s, i + 1, small | (d < s[i] - '0'), nwMask);
    if (res >= mod) {
      res -= mod;
    }
  }
  return res;
}
int solve(const string &s, int i, int small, int mask) {
  if (i == s.size()) {
    return 0;
  }
  int &res = SUM[i][small][mask];
  if (res != -1) {
    return res;
  }
  res = 0;
  for (int d = 0; d < 10; d++) {
    if (!small && d > s[i] - '0') {
      continue;
    }
    int nwMask = mask | (1 << d);
    if (!mask && !d) nwMask = 0;
    if (__builtin_popcount(nwMask) > k) {
      continue;
    }
    res += (d * PW[s.size() - 1 - i] % mod) *
           cnt(s, i + 1, small | (d < s[i] - '0'), nwMask) % mod;
    if (res >= mod) {
      res -= mod;
    }
    res += solve(s, i + 1, small | (d < s[i] - '0'), nwMask);
    if (res >= mod) {
      res -= mod;
    }
  }
  return res;
}
int main() {
  PW[0] = 1;
  for (int i = 1; i < 20; i++) {
    PW[i] = PW[i - 1] * 10 % mod;
  }
  long long int l, r;
  cin >> l >> r >> k;
  string s = str(r);
  memset(SUM, -1, sizeof(SUM));
  memset(C, -1, sizeof(C));
  int ans = solve(s, 0, 0, 0);
  s = str(l - 1);
  memset(SUM, -1, sizeof(SUM));
  memset(C, -1, sizeof(C));
  ans -= solve(s, 0, 0, 0);
  if (ans < 0) {
    ans += mod;
  }
  printf("%d\n", ans);
  return 0;
}
