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;
const int M = 20;
pair<int, int> dp[M][(1 << 10)];
int a[M], K, ten[M];
int Count(int st) {
  int rnt = 0;
  while (st) st &= st - 1, rnt++;
  return rnt;
}
pair<int, int> f(int pos, int st, int lead, int limit) {
  if (pos == -1)
    return Count(st) <= K ? pair<int, int>(0, 1) : pair<int, int>(0, 0);
  if (!limit && dp[pos][st].first != -1) return dp[pos][st];
  int up = limit ? a[pos] : 9;
  pair<int, int> rnt = pair<int, int>(0, 0);
  for (int k = 0; k <= up; k++) {
    pair<int, int> p = f(pos - 1, k == 0 && lead ? 0 : st | (1 << k),
                         k == 0 && lead, limit && k == up);
    rnt.first =
        (rnt.first + p.first + (long long)k * ten[pos] % MOD * p.second % MOD) %
        MOD;
    rnt.second = (rnt.second + p.second) % MOD;
  }
  if (!limit) dp[pos][st] = rnt;
  return rnt;
}
pair<int, int> Solve(long long n) {
  for (int i = 0; i < M; i++, n /= 10) a[i] = n % 10;
  return f(M - 1, 0, 1, 1);
}
int main() {
  for (int i = 0; i < M; i++) {
    ten[i] = i == 0 ? 1 : ten[i - 1] * 10ll % MOD;
    for (int j = 0; j < (1 << 10); j++) dp[i][j].first = -1;
  }
  long long l, r;
  scanf("%I64d%I64d%d", &l, &r, &K);
  pair<int, int> R = Solve(r), L = Solve(l - 1);
  printf("%d\n", (R.first - L.first + MOD) % MOD);
  return 0;
}
