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 mo = 998244353;
int a[20];
int k;
long long ans;
long long ten[20];
pair<long long, long long> f[20][1 << 10][2][2];
pair<long long, long long> dfs(int pos, int mask, bool up, bool zr) {
  if (__builtin_popcount(mask) > k) return {0, 0};
  if (pos == 0) return {1, 0};
  pair<long long, long long> &ret = f[pos][mask][up][zr];
  if (ret.first != -1) return ret;
  ret = {0, 0};
  for (int i = 0; i < 10; i++) {
    if (up && i > a[pos]) continue;
    bool upp = up;
    if (i < a[pos]) upp = 0;
    bool zrr = zr;
    if (i) zrr = 0;
    int maskk = mask;
    if (!zrr) maskk = mask | 1 << i;
    pair<long long, long long> tmp = dfs(pos - 1, maskk, upp, zrr);
    ret.second = (ret.second + i * ten[pos - 1] % mo * tmp.first % mo) % mo;
    ret.first = (ret.first + tmp.first) % mo;
    ret.second = (ret.second + tmp.second) % mo;
  }
  return ret;
}
long long solve(long long first) {
  ten[0] = 1;
  for (int i = 1; i < 20; i++) ten[i] = ten[i - 1] * 10 % mo;
  int n = 0;
  while (first) a[++n] = first % 10, first /= 10;
  memset(f, -1, sizeof(f));
  pair<long long, long long> ans = dfs(n, 0, 1, 1);
  return ans.second;
}
int main() {
  long long l, r;
  cin >> l >> r >> k;
  printf("%lld\n", (solve(r) - solve(l - 1) + mo) % mo);
}
