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 poww[20];
int num[20], len, k;
inline int popcount(long long x) {
  int res = 0;
  while (x) {
    res++;
    x -= x & -x;
  }
  return res;
}
const int mod = 998244353;
pair<long long, long long> che[20][1024][2][2];
pair<long long, long long> dfs(int x, int st, bool lim, bool zero) {
  if (x == 0) {
    return make_pair(0ll, 1ll);
  }
  if (~che[x][st][lim][zero].first) return che[x][st][lim][zero];
  pair<long long, long long> ans = make_pair(0ll, 0ll);
  for (int i = 0; i <= 9; ++i) {
    if (lim && i > num[x]) break;
    bool newzero = zero && i == 0;
    bool newlim = lim && i == num[x];
    int newst = newzero ? st : st | 1 << i;
    if (popcount(newst) > k) continue;
    pair<long long, long long> res = dfs(x - 1, newst, newlim, newzero);
    (ans.first += res.first + res.second * i % mod * poww[x - 1] % mod) %= mod;
    (ans.second += res.second) %= mod;
  }
  return che[x][st][lim][zero] = ans;
}
inline long long f(long long x) {
  memset(che, 0xff, sizeof che);
  len = 0;
  while (x) {
    num[++len] = x % 10;
    x /= 10;
  }
  return dfs(len, 0, 1, 1).first;
}
long long l, r;
int main() {
  poww[0] = 1;
  for (int i = 1; i <= 19; ++i) {
    poww[i] = poww[i - 1] * 10 % mod;
  }
  scanf("%lld%lld%d", &l, &r, &k);
  cout << (f(r) - f(l - 1) + mod) % mod << endl;
}
