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 l, r, k;
long long cnt[20][1 << 10][2];
long long sum[20][1 << 10][2];
long long power[20];
long long calc(string s) {
  memset(cnt, 0, sizeof(cnt));
  memset(sum, 0, sizeof(sum));
  cnt[0][0][1] = 1;
  int len = s.size();
  for (int i = 0; i < len; ++i)
    for (int mask = 0; mask < (1 << 10); ++mask) {
      for (int digit = (i == 0 ? 1 : 0); digit <= 9; ++digit) {
        cnt[i + 1][mask | (1 << digit)][0] += cnt[i][mask][0];
        sum[i + 1][mask | (1 << digit)][0] +=
            (sum[i][mask][0] + digit * power[len - i - 1] * cnt[i][mask][0]) %
            mod;
        if (sum[i + 1][mask | (1 << digit)][0] >= mod)
          sum[i + 1][mask | (1 << digit)][0] -= mod;
      }
      for (int digit = (i == 0 ? 1 : 0); digit < s[i] - '0'; ++digit) {
        cnt[i + 1][mask | (1 << digit)][0] += cnt[i][mask][1];
        sum[i + 1][mask | (1 << digit)][0] +=
            (sum[i][mask][1] + digit * power[len - i - 1] * cnt[i][mask][1]) %
            mod;
        if (sum[i + 1][mask | (1 << digit)][0] >= mod)
          sum[i + 1][mask | (1 << digit)][0] -= mod;
      }
      int digit = s[i] - '0';
      cnt[i + 1][mask | (1 << digit)][1] += cnt[i][mask][1];
      sum[i + 1][mask | (1 << digit)][1] +=
          (sum[i][mask][1] + digit * power[len - i - 1] * cnt[i][mask][1]) %
          mod;
      if (sum[i + 1][mask | (1 << digit)][1] >= mod)
        sum[i + 1][mask | (1 << digit)][1] -= mod;
    }
  long long res = 0;
  for (int mask = 0; mask < (1 << 10); ++mask)
    if (__builtin_popcount(mask) <= k &&
        (cnt[len][mask][0] || cnt[len][mask][1])) {
      res += sum[len][mask][0];
      if (res >= mod) res -= mod;
      res += sum[len][mask][1];
      if (res >= mod) res -= mod;
    }
  return res;
}
long long solve(long long n) {
  int len = to_string(n).size();
  long long res = 0;
  for (int i = 1; i < len; ++i) res += calc(string(i, '9'));
  res += calc(to_string(n));
  return res;
}
int main() {
  power[0] = 1;
  for (int i = 1; i < 20; ++i) power[i] = power[i - 1] * 10;
  cin >> l >> r >> k;
  cout << (solve(r) - solve(l - 1) + mod) % mod;
}
