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 l, r, k, n, pw[20];
string s;
long long int dp[20][1 << 10][2][2];
long long int dp1[20][1 << 10][2][2];
long long int ways(long long int idx, long long int mask, long long int fg,
                   long long int st) {
  long long int bc = __builtin_popcount(mask);
  if (bc > k) return 0;
  if (idx == n)
    return st;
  else if (dp1[idx][mask][fg][st] != -1)
    return dp1[idx][mask][fg][st];
  else {
    long long int val = 0;
    long long int lim = 9;
    if (fg) lim = s[idx] - '0';
    if (st) {
      for (long long int i = 0; i <= lim; i++) {
        val += ways(idx + 1, mask | (1 << i), fg & (i == lim), st);
        val %= 998244353;
      }
    } else {
      for (long long int i = 1; i <= lim; i++) {
        val += ways(idx + 1, mask | (1 << i), fg & (i == lim), 1);
        val %= 998244353;
      }
      val += ways(idx + 1, mask, 0, 0);
      val %= 998244353;
    }
    return dp1[idx][mask][fg][st] = val;
  }
}
long long int solve(long long int idx, long long int mask, long long int fg,
                    long long int st) {
  long long int bc = __builtin_popcount(mask);
  if (bc > k) return 0;
  if (idx == n)
    return 0;
  else if (dp[idx][mask][fg][st] != -1)
    return dp[idx][mask][fg][st];
  else {
    long long int val = 0;
    long long int lim = 9;
    if (fg) lim = s[idx] - '0';
    if (st) {
      for (long long int i = 0; i <= lim; i++) {
        long long int temp =
            ways(idx + 1, mask | (1 << i), fg & (i == lim), st);
        val += (temp * ((i * pw[n - idx - 1]) % 998244353));
        val %= 998244353;
        val += solve(idx + 1, mask | (1 << i), fg & (i == lim), st);
        val %= 998244353;
      }
    } else {
      for (long long int i = 1; i <= lim; i++) {
        long long int temp = ways(idx + 1, mask | (1 << i), fg & (i == lim), 1);
        val += (temp * ((i * pw[n - idx - 1]) % 998244353));
        val %= 998244353;
        val += solve(idx + 1, mask | (1 << i), fg & (i == lim), 1);
        val %= 998244353;
      }
      val += solve(idx + 1, mask, 0, 0);
      val %= 998244353;
    }
    return dp[idx][mask][fg][st] = val;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long int i, j;
  cin >> l >> r >> k;
  pw[0] = 1;
  for (i = 1; i <= 19; i++) pw[i] = (10 * pw[i - 1]) % 998244353;
  memset(dp, -1, sizeof dp);
  memset(dp1, -1, sizeof dp1);
  s = to_string(r);
  n = s.size();
  long long int ans = solve(0, 0, 1, 0);
  l--;
  s = to_string(l);
  n = s.size();
  memset(dp, -1, sizeof dp);
  memset(dp1, -1, sizeof dp1);
  ans -= solve(0, 0, 1, 0);
  ans += 998244353;
  ans %= 998244353;
  cout << ans;
  return 0;
}
