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;
int k, mod = 998244353, pw[19];
array<int, 2> dp[19][1 << 10][2];
string s;
int add(int a, int b) { return (a + 0LL + b) % mod; }
int mul(int a, int b) { return (a * 1LL * b) % mod; }
array<int, 2> calc(int i = 0, int mask = 0, bool eq = 0) {
  if (__builtin_popcount(mask) > k) {
    return {0};
  }
  if (i == s.size()) {
    return {0, 1};
  }
  array<int, 2> &ret = dp[i][mask][eq];
  if (~ret[0]) {
    return ret;
  }
  ret = {0};
  for (int c = 0; c <= (eq ? 9 : s[i] - '0'); ++c) {
    array<int, 2> res =
        calc(i + 1, !mask && !c ? 0 : mask | (1 << c), eq | (c < s[i] - '0'));
    ret[0] = add(ret[0], res[0]);
    ret[1] = add(ret[1], res[1]);
    ret[0] = add(ret[0], mul(mul(c, pw[s.size() - i - 1]), res[1]));
  }
  return ret;
}
int solve(long long x) {
  s = to_string(x);
  for (int i = 0; i < 19; ++i)
    for (int j = 0; j < 1 << 10; ++j)
      for (int k = 0; k < 2; ++k) dp[i][j][k] = {-1, -1};
  return calc()[0];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  pw[0] = 1;
  for (int i = 1; i < 19; ++i) {
    pw[i] = mul(pw[i - 1], 10);
  }
  long long l, r;
  cin >> l >> r >> k;
  cout << add((solve(r) - 0LL - solve(l - 1)) % mod, mod) << "\n";
  return 0;
}
