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 INF = 0x7f7f7f7f7f7f;
const long long mo = 998244353;
long long k, l, cnt, r, a[20], pw[20];
pair<long long, long long> dp[20][(1 << 10) + 5];
pair<long long, long long> dfs(long long pos, long long S, bool lead,
                               bool lim) {
  if (!~pos) return pair<long long, long long>(__builtin_popcount(S) <= k, 0);
  if (!lead && !lim && ~dp[pos][S].first) return dp[pos][S];
  long long up = lim ? a[pos] : 9;
  pair<long long, long long> ans = make_pair(0, 0);
  for (long long i = 0; i <= up; i++) {
    long long S1 = (lead && i == 0) ? S : (S | (1 << i));
    pair<long long, long long> nxt =
        dfs(pos - 1, S1, lead && i == 0, lim && i == up);
    ans = pair<long long, long long>(
        (ans.first + nxt.first) % mo,
        (nxt.first * pw[pos] % mo * i % mo + nxt.second + ans.second) % mo);
  }
  if (!lim && !lead) dp[pos][S] = ans;
  return ans;
}
long long calc(long long x) {
  memset(a, 0, sizeof(a));
  cnt = 0;
  while (x) a[cnt++] = x % 10, x /= 10;
  for (long long i = 0; i <= 18; i++)
    for (long long j = 0; j < (1 << 10); j++)
      dp[i][j].first = dp[i][j].second = -1;
  return dfs(cnt - 1, 0, 1, 1).second;
}
signed main() {
  pw[0] = 1;
  for (long long i = 1; i <= 18; i++) pw[i] = pw[i - 1] * 10ll % mo;
  scanf("%lld %lld %lld", &l, &r, &k);
  printf("%lld\n", (calc(r) - calc(l - 1) + mo) % mo);
  return 0;
}
